Which functions have the same arc length to the origin at every point as the *signed* curvature at the point? Which functions $y\left( x \right)$ have the same arc length $s$ to the origin at every point as the signed curvature $\kappa$ at the point?
We know the formulas for the arc length $s$
$$
s\left( a, b, \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right) = \int_{a}^{b} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x
$$
and signed curvature $\kappa$
$$
\kappa\left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x}, \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}} \right) = \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}} \text{.}\\
$$
Since we are talking about the length of the function $y\left( x \right)$ from the origin to every point, here $a = 0$ and $b = x$, giving us a nonlinear, higher-order ordinal differential equation:
$$
\begin{align*}
s\left( 0, x, \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right) &= \kappa\left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x}, \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}} \right)\\
\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}}\\
\end{align*}
$$
Now comes the hard part: Solving the ODE for y(x).
(From here only my failed attempts follow...)

My first attempt at solving the ODE
Since I can't think of a specific solution, I would try a bit of reshaping first.
$$
\begin{align*}
\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}}\\
\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{3 \cdot \frac{1}{2}}}\\
\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \right)^{3}} \quad\mid\quad u\left( x \right) := \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}}\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{u\left( x \right)^{3}} \quad\mid\quad \cdot u\left( x \right)^{3} \tag{1.}\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x \cdot u\left( x \right)^{3} &= \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}\\
\\
u\left( x \right) &= \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \quad\mid\quad \left(  \right)^{2}\\
u\left( x \right)^{2} &= 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \quad\mid\quad -1\\
u\left( x \right)^{2} - 1 &= \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \quad\mid\quad \sqrt{}\\
\sqrt{u\left( x \right)^{2} - 1} &= \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \quad\mid\quad \frac{\operatorname{d}}{\operatorname{d}x}\\
\frac{\frac{\operatorname{d}u\left( x \right)}{\operatorname{d}x}}{2 \cdot \sqrt{u\left( x \right) - 1}} &= \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}} \tag{2.}\\
\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x \cdot u\left( x \right)^{3} &= \frac{\frac{\operatorname{d}u\left( x \right)}{\operatorname{d}x}}{2 \cdot \sqrt{u\left( x \right) - 1}}\\
\end{align*}
$$
And I'm stuck aigan.
My second attempt at solving the ODE
Next, I would try to transform the higher-order nonlinear ODE into a third-order nonlinear ODE, because maybe that will help. Since the substitution simplified the equation in my opinion, I use it again.
$$
\begin{align*}
\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}} \quad\mid\quad \text{useing } \left( 1. \right) \text{ and } \left( 2. \right)\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x &= \frac{\frac{\operatorname{d}u\left( x \right)}{\operatorname{d}x}}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1}} \quad\mid\quad \frac{\operatorname{d}}{\operatorname{d}x}\\
\\
u\left( x \right) &= \dots
\end{align*}
$$
Time to give up... Since it looks harder now.

My numerical attempt
We know from calculus $\frac{\operatorname{d}f\left( x \right)}{\operatorname{d}x} = \frac{f\left( x + h \right) - f\left( x \right)}{h} + \mathcal{O}\left( h \right)$ or $\frac{\operatorname{d}f\left( x \right)}{\operatorname{d}x} \approx \frac{f\left( x + h \right) - f\left( x \right)}{h}, \text{for small } \left| h \right|$ (Euler method) aka:
$$
\begin{align*}
\int_{0}^{x} \sqrt{1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2}} \operatorname{d}x &= \frac{\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}}}{\left( 1 + \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} \right)^{\frac{3}{2}}} \quad\mid\quad \text{useing } \left( 1. \right) \text{ and } \left( 2. \right)\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x &= \frac{\frac{\operatorname{d}u\left( x \right)}{\operatorname{d}x}}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1}}\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x &\approx \frac{\frac{u\left( x + h \right) - u\left( x \right)}{h}}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1}}\\
\int_{0}^{x} u\left( x \right) \operatorname{d}x &\approx \frac{u\left( x + h \right) - u\left( x \right)}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h} \quad\mid\quad \frac{\operatorname{d}}{\operatorname{d}x}\\
u\left( x \right) &\approx \frac{\operatorname{d}}{\operatorname{d}x} \frac{u\left( x + h \right) - u\left( x \right)}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h}\\
u\left( x \right) &\approx \frac{\frac{u\left( x + 2 \cdot h \right) - u\left( x + h \right)}{2 \cdot u\left( x + h \right)^{3} \cdot \sqrt{u\left( x + h \right) - 1} \cdot h} - \frac{u\left( x + h \right) - u\left( x \right)}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h}}{h}\\
u\left( x \right) &\approx \frac{u\left( x + 2 \cdot h \right) - u\left( x + h \right)}{2 \cdot u\left( x + h \right)^{3} \cdot \sqrt{u\left( x + h \right) - 1} \cdot h^{2}} - \frac{u\left( x + h \right) - u\left( x \right)}{2 \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2}}\\
u\left( x \right) &\approx \frac{u\left( x + 2 \cdot h \right) \cdot u\left( x \right)^{3} \cdot u\left( x \right)^{3} - u\left( x + h \right) \cdot u\left( x \right)^{3}}{2 \cdot u\left( x + h \right)^{3} \cdot \sqrt{u\left( x + h \right) - 1} \cdot h^{2}} - \frac{u\left( x + h \right) \cdot u\left( x + h \right)^{3} - u\left( x \right) \cdot u\left( x + h \right)^{3}}{2 \cdot u\left( x + h \right)^{3} \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2}}\\
u\left( x \right) &\approx \frac{u\left( x + 2 \cdot h \right) \cdot u\left( x \right)^{3} \cdot u\left( x \right)^{3} - u\left( x + h \right) \cdot u\left( x \right)^{3} - \left( u\left( x + h \right) \cdot u\left( x + h \right)^{3} - u\left( x \right) \cdot u\left( x + h \right)^{3} \right)}{2 \cdot u\left( x + h \right)^{3} \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2}} \quad\mid\quad \cdot \left( 2 \cdot u\left( x + h \right)^{3} \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2} \right)\\
u\left( x \right) \cdot 2 \cdot u\left( x + h \right)^{3} \cdot u\left( x \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2} &\approx u\left( x + 2 \cdot h \right) \cdot u\left( x \right)^{3} \cdot u\left( x \right)^{3} - u\left( x + h \right) \cdot u\left( x \right)^{3} - \left( u\left( x + h \right) \cdot u\left( x + h \right)^{3} - u\left( x \right) \cdot u\left( x + h \right)^{3} \right)\\
u\left( x \right)^{4} \cdot 2 \cdot u\left( x + h \right)^{3} \cdot \sqrt{u\left( x \right) - 1} \cdot h^{2} &\approx u\left( x + 2 \cdot h \right) \cdot u\left( x \right)^{6} - u\left( x + h \right) \cdot u\left( x \right)^{3} - u\left( x + h \right)^{4} + u\left( x \right) \cdot u\left( x + h \right)^{3}\\
\end{align*}
$$
Time to give up, since this dosn't look solvable for me...

I am grateful for every help, correction and suggestion.

First, notice that the defining condition determines an Euler spiral.
Per OP's comment, here's how to construct an explicit parameterization of an (essentially general) curve $\gamma$ satisfying the condition. In the end we'll exploit the construction to describe a function $y(x)$ satisfying the condition.
The Frenet equations in $2$ dimensions, satisfied by the curvature $\kappa(s)$, tangent vector ${\bf T}(s)$, and normal vector ${\bf N}(s)$, are
$$\pmatrix{{\bf T}' & {\bf N}'} = \pmatrix{{\bf T} & {\bf N}} \pmatrix{\cdot&-\kappa\\\kappa&\cdot} .$$ As usual, $s$ is the arc length parameter.
In our case, $\kappa(s) = s$, so our differential equation becomes
$$\pmatrix{{\bf T}' & {\bf N}'} = \pmatrix{{\bf T} & {\bf N}} s \pmatrix{\cdot&-1\\1&\cdot},$$
and rearranging gives
$$\pmatrix{{\bf T} & {\bf N}}^{-1}\pmatrix{{\bf T}' & {\bf N}'} = s \pmatrix{\cdot&-1\\1&\cdot} ,$$
so
\begin{align*}
\pmatrix{{\bf T} & {\bf N}}
&= \pmatrix{{\bf T}(0) & {\bf N}(0)} \exp \int_0^s \left[u\pmatrix{\cdot&-1\\1&\cdot}\right] du\\
&= \pmatrix{{\bf T}(0) & {\bf N}(0)} \exp \left[\pmatrix{\cdot&-1\\1&\cdot} \int_0^s u \,du\right] \\
&= \pmatrix{{\bf T}(0) & {\bf N}(0)} \exp \left[\frac{1}{2} s^2 \pmatrix{\cdot&-1\\1&\cdot}\right] \\
&=\pmatrix{{\bf T}(0) & {\bf N}(0)} \pmatrix{\cos \frac{s^2}{2} & \ast \\ \sin \frac{s^2}{2} & \ast}.
\end{align*}
If we denote ${\bf T}(0) = (\cos \theta, \sin \theta)$, then
$${\bf T} = \pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta} \pmatrix{\cos \frac{s^2}{2} \\ \sin \frac{s^2}{2}}.$$ Integrating and setting $\gamma(0) = 0$ gives
$$\gamma(s) = \int_0^s {\bf T}'(u) \,du = \pmatrix{\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta} \pmatrix{\int_0^s \cos \frac{u^2}{2} \,du \\  \int_0^s \sin \frac{u^2}{2} \,du} .$$ So, up to rotation, there is a unique curve satisfying $\gamma(0) = 0$. Taking $\theta = 0$ to choose a representative we find that
$$\color{#df0000}{\boxed{\gamma(s) = \pmatrix{\int_0^s \cos \frac{u^2}{2} \,du \\ \int_0^s \sin \frac{u^2}{2} \,du} = \pmatrix{\sqrt{2} C\left(\frac{s}{\sqrt 2}\right) \\ \sqrt{2} S\left(\frac{s}{\sqrt 2}\right)}}},$$ where $C$ and $S$ are the Fresnel integral functions.
The solution curve $\gamma$, $s \in [-32, 32]$.

The Fresnel cosine function $C$ is injective in a neighborhood of $x = 0$, so it admits a local inverse, $C^{-1}$, in which case a solution $y(x)$ to the original o.d.e. is $$\color{#df0000}{\boxed{y(x) = \sqrt{2} S\left(C^{-1}\left(\frac{x}{\sqrt{2}}\right)\right)}} ,$$ in agreement with achille hui's comment.
The function $y(x)$ with maximum domain.

We can verify directly that the truncation of the power series of $y$ agrees with that in my other answer:
$$y(x) = \frac16 x^3 + \frac1{105} x^7 + \frac{293}{237600} x^{11} + \cdots .$$
