Since in retrospect I realise this question can be taken in several different ways, I am posting another reply myself, which combines some of the best answers so far and could hopefully help somebody else understand why this obviously wrong physical reasoning occurs mathematically.
As is clear at this point, the original issue was bad notation and there are at least four different ways in which this question can be understood (and resolved). From now on I am going to refer to the current position as $\vec{x}_0$ and the trajectory as a function of this starting position and time as $\vec{x}(t)$, such that $\vec{x}(0)=\vec{x}_0$, in order to avoid confusion. In all cases the other initial condition $\vec{\dot{x}}(0)=\vec{v}_0$ will not be given much consideration, but we still have to be aware of it.
1. First scenario: as correctly pointed out, <PERSON>'s second law does not imply that the force comes from a potential and/or is a function of the current position. In this case the force is simply a definition and the question doesn't hold anymore, since the force does not have to be a function of the trajectory:
$$\vec{F}(t)=m\vec{\ddot{x}}(t)$$
1. Second scenario: the force comes from a potential, such that $\vec{F}(\vec{x}(t))=-\left.\frac{\partial U(\vec{x}')}{\partial \vec{x}'}\right|_{\vec{x}'=\vec{x}(t)}$ and we make a coordinate transformation of the whole trajectory $\vec{x}(t)=s\vec{y}(t)$ for some scaling factor $s$. In this case:$$\vec{F}(\vec{x}(t))=m\vec{\ddot{x}}(t) \Leftrightarrow \vec{F}(s\vec{y}(t))=ms\vec{\ddot{y}}(t), \qquad\vec{F}(\vec{x}(t))=\vec{F}(s\vec{y}(t))$$ This is a simple change in variables and there is not much else to be found here - certainly no contradictions anyway.
2. Third scenario: the force comes from a potential and we rescale the initial condition of the differential equation, so that $\vec{\tilde{x}}(0)=s\vec{x}_0$. In general, this trajectory will be completely different to $\vec{x}(t)$ and the only way $\vec{\tilde{x}}(t)=s\vec{x}(t)$ is if the force corresponds to a quadratic potential.
3. Fourth scenario: the force comes from a potential and we rescale the solution $x(t)$ of the differential equation, such that $\vec{\tilde{x}}(t)=s\vec{x}(t)$. 766
In general, this rescaled function no longer satisfies the original differential equation and nothing else can be said. The only way the new rescaled trajectory satisfies the original differential equation is if we had a harmonic potential.
Note that the difference between 3. and 4. is subtle: the new trajectory $\vec{\tilde{x}}(t)$ in 3. satisfies <PERSON>'s second law but its shape is in general different to $\vec{x}(t)$, while the $\vec{\tilde{x}}(t)$ in 4. doesn't in general satisfy <PERSON>'s second law (and doesn't have to) and its shape is the same as $\vec{x}(t)$. In both 3. and 4. the trajectories can only degenerate for a harmonic potential (although one might have to fiddle with the velocity initial condition to achieve that), which is what my original post "proved", although it certainly isn't clear whether I proved 3. or 4. (possibly neither) due to my bad notation! 766