Let $\alpha:ֿ\mathbb{S}^1 \to \mathbb{R}^2$ be a $C^2$ curve satisfying $|\dot \alpha|=1$. Define $b=\langle \alpha,\dot \alpha\rangle$ and assume that $$ b(\theta+c_1)-b(\theta)=c_2 \tag{1} $$ holds for every $\theta$, where $c_1,c_2$ are constants. Does $c_2=0$?
(If $c_2=0$, then a "rotation" by $c_1$ is a sort of "symmetry" of $\alpha$.)
Differentiating equation $(1)$ we get $\dot b(\theta+c_1)=\dot b(\theta)$. Since $\dot b=1+\langle \alpha,\ddot \alpha\rangle$, this is equivalent to $$ \langle \alpha,\ddot \alpha\rangle(\theta)=\langle \alpha,\ddot \alpha\rangle(\theta+c_1). $$
I am not sure how to proceed from here.
I think that the answer is positive:
Write $f(t)=|\alpha(t)|^2$. Then $f'=2b$.
Our assumption is equivalent to $$ f'(t+c_1)-f'(t)=2c_2. $$ Since $f$ is $2\pi$-periodic, we have $$ 0=\int_{0}^{2\pi} f'(t)dt=\int_{0}^{2\pi} f'(t+c_1)dt. $$ Since $$ \int_{0}^{2\pi} f'(t+c_1)dt-\int_{0}^{2\pi} f'(t)dt=4\pi c_2, $$ this forces $c_2=0$.