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.

I can't see a closed-form solution for an explicit function $y(x)$, but we can make some progress.
Differentiating both sides of the equation and rearranging gives the $3$rd order o.d.e.
$$(1 + (y')^2)^3 = y'''(x) \left[(y')^2 + 1\right] - 3 (y'')^2 y' ,$$ and writing $u := y'$ reduces it to a $2$nd order equation,
$$(1 + u^2)^3 = u'' (1 + u^2) - 3 (u')^2 u .$$
Swapping the independent and dependent variables, i.e., regarding $x$ as a function of $u$, yields the equation
$$x''(u) = -(1 + u^2)^2 x'(u)^3 - \frac{3 u x'(u)}{1 + u^2}.$$ We may now reduce to a $1$st order equation in $z := x'$:
$$z' = -(1 + u^2)^2 z^3 - \frac{3 u z}{1 + u^2} .$$
Substituting $z(u) = \frac{1}{(1 + u^2)^{3 / 2} \sqrt{w(u)}}$ transforms the equation to
$$w'(u) = \frac{2}{u^2 + 1},$$
or
$$w(u) = 2 \arctan u + C.$$
Backsubstituting yields the generic solutions
$$z(u) = \pm \frac{1}{(u^2 + 1)^{3 / 2} (\sqrt{2 \arctan u + C})}$$
and thus
$$x(u) = \pm \int_a^u \frac{dt}{(t^2 + 1)^{3 / 2} (\sqrt{2 \arctan t + C})} .$$ The integral on the right-hand side likely has no closed form, but the inverse of this $x(u)$ is $u(x) = y'(x)$, and integrating again gives $y(x)$.
Incidentally, we can compute the truncation of a series solution: Setting $y(0) = 0$, $y'(0) = a$, $y''(0) = b$ gives
$$y(x) = a x + \frac{1}{2} b x^2 + \frac{1}{6} \left[(1 + a^2)^2 + \frac{3 a b^2}{1 + a^2}\right] x^3 + \cdots .$$
Taking $a = b = 0$ yields the solution with the highest order of tangency to the $x$-axis:
$$y(x) = \frac16 x^3 + \frac1{105} x^7 + \frac{293}{237600} x^{11} + \cdots .$$
