Let $\alpha,\beta:[0,1]\to\mathbb{R}^2$ be two smooth curves satisfying $|\alpha(t)| = |\beta(t)|$ and $|\dot{\alpha}(t)| = |\dot{\beta}(t)|$ for all $t\in[0,1]$. That is, $\alpha$ and $\beta$ have the same speed and are at the same distance to the origin.
Does there exist a rotation matrix $Q$ (with possibly negative determinant) such that $\alpha(t) = Q\beta(t)$?
It seems that the answer is yes. The speed $|\dot{\alpha}|$ should be enough to tell how quickly the curve is going around the circle. This seems to be enough to recover the angular data of the curve, but I do not know of a good way to make this precise.
Unfortunately it is not true. Let
$$ f(x) = \begin{cases} e^{-1/x}, &x >0, \\ -e^{-1/x}, & x<0, \\ 0 , & x=0. \end{cases}.$$ Then $\alpha (t) = (f(t), 0)$ and $$\beta (t) =\begin{cases} (f(t) , 0) & t \le 0, \\(0,f(t)) & t >0\end{cases}$$ satisfy your condition, but there is no rotation sending the straight line $\alpha$ to a corner $\beta$.
On the other hand, your assertion is true when you assume that the curves do not passes through the origin. To see this, let \begin{align} \alpha (t) &= r(t) e^{i\theta_\alpha (t)}, \\ \beta (t) &= r(t) e^{i\theta_\beta (t)}, \end{align}
(note that we have the same $r$ since $|\alpha (t)| = |\beta (t)|$. Differentiating gives
\begin{align} \dot\alpha (t) &= \dot r(t) e^{i\theta_\alpha (t)} + i\dot\theta_\alpha r(t)e^ {i\theta_\alpha(t)}, \\ \dot\beta (t) &= \dot r(t) e^{i\theta_\beta (t)} + i\dot\theta_\beta r(t)e^ {i\theta_\beta(t)}, \end{align} and
\begin{align} |\dot\alpha (t)| &= |\dot r(t)| +|\dot\theta_\alpha| |r(t)|, \\ |\dot\beta (t)| &= |\dot r(t)| +|\dot\theta_\beta| |r(t)|. \end{align}
Since $r(t) \neq 0$ for all $t$, one concludes that $\dot\theta_\alpha (t) = \pm \dot\theta_\beta (t)$, or
$$ \theta_\alpha (t) = \pm \theta_\beta (t) + C.$$
This implies the existence of such $Q$.