A question about curves in $\mathbb{R}^2$

Ok let me try it again.

First of all, suppose that $\alpha$ is parametrized by arc length.

Write $$\alpha(s)-\alpha(t)=\alpha'(t)(s-t)+\alpha''(t)\frac{(s-t)^{2}}{2}+o(|s-t|)$$

Then

\begin{eqnarray} \|\alpha(s)-\alpha(t)\|^{2} &=& \|\alpha'(t)(s-t)+\alpha''(t)\frac{(s-t)^{2}}{2}+o(|s-t|)\|^{2} \nonumber \\ &=& \|\alpha'(t)(s-t)+\alpha''(t)\frac{(s-t)^{2}}{2}\|^{2}+2\langle\alpha'(t)(s-t)+\alpha''(t)\frac{(s-t)^{2}}{2},o(|s-t|)\rangle \nonumber \\ &&+\|o(|s-t|)\|^{2} \nonumber \\ &=& |s-t|^{2}+k(t)\frac{|s-t|^{4}}{4}+o(|s-t|) \end{eqnarray}

So, as you can see, if $k(t)$ is not constant $\|\alpha(s)-\alpha(t)\|$ depends on $k(t)$. Hence the only possibilities are the circle or the line. But an straightforward calculation shows that for the circle and the line the statement is true.

You do not need differentiability. Below is a proof that works when $\alpha$ is only continuous.

We may assume that $I$ starts with $0$ : $I=[0,M]$ for some $M$. Suppose that the image of $\alpha$ is not contained in a line. Then there are $x_1<x_2<x_3$ such that $\alpha(x_1),\alpha(x_2),\alpha(x_3)$ are not collinear. Let $t\in ]0,M-x_3[$. Let now $x=x_4$ be an arbitrary number in $[0,M-t]$. Consider the two families of four points $A_i=\alpha(x_i) (1 \leq i \leq 4)$ and $B_i=\alpha(x_i+t) (1 \leq i \leq 4)$. They are isometric by hypothesis : $d(A_i,A_j)=d(B_i,B_j)$.

By this older StackOverflow question, there is an isometry $\gamma(x,t)$ of the plane sending $A_i$ to $B_i$. Since $\gamma(t)$ is already uniquely determined by its image on $\alpha(x_1),\alpha(x_2),\alpha(x_3)$, we see that $\gamma(x,t)=\gamma(t)$ is in fact independent of $x$, and we can explicitly write out the coefficients of the matrix of $\gamma(t)$ in terms of the coordinates of $\alpha(x_1),\alpha(x_2),\alpha(x_3)$ and $\alpha(x_1+t),\alpha(x_2+t),\alpha(x_3+t)$. So $\gamma$ will be continuous if $\alpha$ is. Unicity also ensures that $\gamma$ is a homomorphism (where it is defined) : $\gamma(s+t)=\gamma(s)\gamma(t)$ when $s,t$ and $s+t$ are all in $I$.

Now ${\sf det}(\gamma(t))=1$ or $-1$, and by continuity this determinant is always $1$ or always $-1$. Since $\gamma(0)$ is the identity, we see that $\gamma$ is always an (affine) rotation. We have $\gamma(s)\gamma(t)=\gamma(s+t)=\gamma(t)\gamma(s)$. We see that $\gamma(s)$ and $\gamma(t)$ commute, so they must share the same center point. We deduce that the center $\Omega$ of $\gamma(t)$ is independent of $t$. Now for any $s<t$, $\alpha(t)=\gamma(t-s)\alpha(s)$, so $\alpha(t)$ and $\alpha(s)$ are located at the same distance from $\Omega$. So $\alpha$ walks on a circle of cnter $\Omega$, qed.