This question is a follow-up of this one.
Let $\alpha:\mathbb{S}^1\to \mathbb{R}^2$ be a $C^2$ curve satisfying $|\dot \alpha|=1$, and define $s= \alpha \times \dot \alpha$, where $\times$ denotes the cross-product of vectors in $\mathbb{R}^3$. (I think of $\alpha, \dot \alpha$ as lying in the plane $z=0$ inside $\mathbb{R}^3$).
Suppose that $s(\theta)$ is nowhere locally constant, has no zeroes, and that $$ s(\theta+c)=s(\theta) \tag{1} $$ holds for every $\theta$, where $c$ is a fixed constant.
Question: Does there exist a rotation $R \in \text{SO}(2) $ such that $\alpha(\theta+c)=R\alpha(\theta)$ holds for every $\theta$?
Comments:
Whenever such an $R$ exists, then, clearly equation $(1)$ holds. Indeed, $$ s(\theta+c)= \alpha(\theta+c) \times \dot \alpha(\theta+c)= R\alpha(\theta) \times R\dot \alpha(\theta) =R\big(\alpha(\theta) \times \dot \alpha(\theta) \big)=$$ $$\alpha(\theta) \times \dot \alpha(\theta)=s(\theta), $$ where we have used the fact that, since $R \in \text{SO}(2) $ , it fixes the $z$-axis, and in particular, fixes $\alpha(\theta) \times \dot \alpha(\theta) \in \text{span}(e_3)$.
It suffices to prove the following a-priori weaker looking conclusion:
There exists a $C^1$ mapping $R:\mathbb{S}^1 \to \text{SO}(2)$ such that $$ \alpha(\theta+c)=R(\theta)\alpha(\theta) \tag{2} $$ holds for every $\theta$.
Indeed, differentiating this, one gets $$ \dot \alpha(\theta+c)=\dot R(\theta)\alpha(\theta)+R(\theta)\dot \alpha(\theta), $$ so $$ s(\theta+c)= \alpha(\theta+c) \times \dot \alpha(\theta+c)= R(\theta)\alpha(\theta) \times \dot R(\theta)\alpha(\theta)+R(\theta)\dot \alpha(\theta)= $$ $$ s(\theta)+R(\theta)\alpha(\theta) \times \dot R(\theta)\alpha(\theta). $$ Thus $s(\theta+c)=s(\theta)$ implies that $$ R(\theta)\alpha(\theta) \times \dot R(\theta)\alpha(\theta)=0. $$ Assume by contradiction that $\dot R(\theta) \neq 0$. Then, since $\alpha(\theta) \neq 0$, $R(\theta)\alpha(\theta), \dot R(\theta)\alpha(\theta)$ are both non-zero parallel vectors in $\mathbb{R}^2$.
Writing in complex numbers notation, $R(\theta)=e^{i\phi(\theta)}$, $\dot R(\theta)=i\phi'(\theta)R(\theta)$, so $$ \dot R(\theta)\alpha(\theta)= i\phi'(\theta)R(\theta)\alpha(\theta). $$ Since $R(\theta)\alpha(\theta), \dot R(\theta)\alpha(\theta)$ are parallel, they should be proportional to each other with a real ratio.
Thus we must have $\phi'(\theta)=0$, i.e. $\dot R(\theta) = 0$.
Some musings that unfortunately don't reach all the way home.
Assume that $s$ is positive everywhere -- this does not lose significant generality because otherwise it's negative everywhere, and then mirroring everything will get us back to the positive case.
The main insight I have to offer is that we can reduce this to a one-dimensional problem by writing $\alpha$ in polar coordinates: $$ \alpha(\theta) = r(\theta)e^{i\varphi(\theta)} $$ [The function $g$ in my previous answer was essentially $r$ as a function of $\varphi$ instead of as a function of $\theta$, but for the following argument we need to keep everything parameterized by arc length.]
The condition $|\dot \alpha|=1$ now becomes $$ \tag{1} \dot r^2 + (r\dot\varphi)^2 = 1 \implies \dot\varphi = \frac{\sqrt{1-\dot r^2}}r $$ so we can reconstruct $\varphi$ from $r$ up to a constant of integration, and mostly care only about the behavior of $r$. By geometric considerations we have $$ \tag{2} s = r\cdot r\dot\varphi = r\sqrt{1-\dot r^2} $$ which also tells us that $s\le r$ everywhere.
Finally, (2) can be rearranged to $$ \tag{3} \dot r = \pm\sqrt{1-s^2/r^2} $$ so $\dot r=0$ only when $s=r$. But when that is the case we must have $\dot s=0$ too, or $s$ would be larger than $r$ on one of the sides which makes (2) and (3) impossible.
Since $\alpha$ is a closed curve, $r(\theta)$ has a maximum $M$ -- without loss of generality we can assume that it is attained at $\theta=0$, so $r(0)=M$ is maximal, $r'(0)=0$ and therefore $s(0)=M$ too. We now by assumption also have $s(nc)=M$ for every integer $n$. But then $$ M = s(nc) \le r(nc) \le M $$ so $r(nc)=M$ too.
The question now essentially becomes: Can there be any $s:[0,c]\to\mathbb R_+$ where there are two different ways to reconstruct $r$ on the interval $[0,c]$ such that $r(0)=r(c)=M$? If this not possible, then the answer to your question is "yes". But if we could find an $s$ that matches two different $r$s in this way, then we can probably use it to construct a counterexample (with some appropriate footwork to get the total increase in $\varphi$ over a period to be a rational multiple of $2\pi$ and then scale the solution down so the total arc length is an appropriate small integer fraction of $2\pi$).
Let's therefore assume, hypothetically, that we have $r_1 \ne r_2$ that both give rise to the same $s$. To be concrete about the difference, suppose that for some $b\in(0,c)$ we have $r_1(b) > r_2(b)$.
To begin with, if $r_1>r_2$ on a certain interval of $\theta$, then it is impossible for $\dot r_1$ to change sign in that interval. Namely, if $\dot r_1=0$, then (2) gives $r_1=s$, but that would violate the general fact that $s \le r_2$ everywhere.
We can further assume that $r_1'(b)<0$. (We have just seen that $r_1'(b)=0$ is impossible, and if $r_1(b)>0$ we can just flip the sign of $\theta$ everywhere and start over).
Now, if we move $\theta$ back from $b$ towards $0$ there will be a first time when $r_1(\theta)=r_2(\theta)$. Let's call that time $d$: $$ d = \sup\{ \theta \in (0,b) \mid r_1(\theta) = r_2(\theta) \} $$ By continuity it is clear that $r_1(d)=r_2(d)$.
Since $r_1>r_2$ on the interval $(d,b)$ and $r_1$ is decreasing around $d$, $r_1$ is strictly decreasing on that entire interval, by the argument a few paragraphs above.
Now it must be that $r_1'(d)=r_2'(d)$. If they were different, then (3) tells us that one derivative is positive and the other is negative -- but that contradicts the fact that the larger of $r_1$ and $r_2$ immediately decreases. Furthermore, the common value of $r_1'(d)$ and $r_2'(d)$ must be $0$. If it is nonzero, then (3) behaves sufficiently nicely in a neighborhood of $(d, r_i(d))$ that the Picard-Lindelöf theorem says $r_1$ and $r_2$ cannot actually split at that point.
Summing up the conclusions so far: We can only have $r_1\ne r_2$ if there is a point $d$ with:
At this point my rigor unfortunately dries up.
We know from one of your earlier questions one particular way for curves with the same $s$ to split, namely if one of them continues at constant $r$ (that is, a circular arc) whereas the other veers off in a straight line. Intuivitely if we assume that the features of this split generalize to cases where $s$ is not constant, it seems unlikely that it's possible for both the two curves to continue inside the circular arc. Instead I would expect that if $\ddot s(d)>0$ the curves that split must both lie between the circular arc and the straight line (both $r_i$s too large for the above facts), and if $\ddot s(d)<0$ one of them must lie outside the straight line, which corresponds to an increasing $r_1$. Both of those would contradict what I've already derived.
But I can't prove that to be the case. Perhaps one of the answers to Must a $𝐶^1$ curve with constant angular momentum alternate between a straight line or a circular arc? can be adapted.