When playing around with closed planar curves centered at the origin such as ellipses and circles in "standard" parametrisation (i.e. $(a \cos(t), b \sin(t))$ and period $2 \pi$) I noticed that they are their own derivatives. So I asked myself for which other closed curves this holds. For a curve like $(c, 2 \pi)$, where $c(t) := (\cos(t), \sin(2t))$, the derivative $c'$ obviously does not coincide with $c$. But can we reparametrise $c$ (in a way that is its speed towards this bumps is slowed down) such that the derivative coincides with $c$?
Another example: The derivative of unit circle in standard parametrisation coincides with the unit circle. But if we reparametrise the unit circle as $(\cos(t \cdot e^{t - 2 \pi}), \sin(t \cdot e^{t - 2 \pi}))$ with still period $2 \pi$, the derivative doens't coincide.
Another factor to consider is translated versions of curves. If a circle centered at the origin is translated to another position, its derivative will be centered at the origin.
Thus my question is:
Let $(c,p)$ be a closed planar curve. Does there exists a reparametrisation of $c$ such that $c'$ and $c$ look the same modulo translation?
Definition 1. A closed parametrised curve is a pair $(c, p)$ where $c: \mathbb R \to \mathbb R^n$ is parametrised curve with period $p$, i.e. $c(t+p)=c(t)$ holds for all $t \in \mathbb R$.
Definition 2. A closed curve is an equivalence class of closed parametrised curves, where $(c,p) \sim (d,q)$ if and only if there exists a bijective smooth map $\phi: \mathbb R \to \mathbb R$ such that $d = c \circ \phi$ and $\phi'(t) > 0$ and $\phi(t + p) = \phi(t) + q$ hold for all for all $t \in \mathbb R$
For a simple closed strictly convex regular curve this is true. Assume that $(c,p)$ is positively oriented (for negatively oriented curves the argument is similar). Translate $(c, p)$ so that the origin is in the interior. Now for every $t$ the ray from the origin in the direction $c'(t)$ will intersect the curve at $v(t)=u(t) c'(t)$ for some $u(t)>0$. This is the geometric key point - we just need to reparametrize so that at $c(t)$ the tangent becomes $v(t)$.
In formulas, let $\phi$ be defined by $\phi(0)=0$ and $\phi'(\tau)=u(\phi(\tau))$ (i.e. the IVP solution for the corresponding ODE) . Then $c(\phi(\tau))'=\phi'(\tau) c'(\phi(\tau))=u(\phi(\tau))c'(\phi(\tau))=v(\phi(\tau))$. As $\tau$ goes from $0$ to $\phi^{-1}(p)$ (the period of $d=c\cdot \phi$) the point $v(\phi(\tau))$ goes around the image of $c$ once. Thus $d'$ has the same image as $c$ i.e. same image as $d$.
On the other hand for the curve $(\cos t, \sin 2t)$ this is impossible. The curve's tangent has a rotation number of $0$. At the same time the winding number of the curve around any point in the plane is either $\pm 1$ (when the point is inside one of the loops) or $0$ (when it is in the outer region). This implies that if $c'=c$ after a translation then the origin must be outside $c$. But then there exists a vector $v$ such vectors $c(t)$ will never be positively proportional to $v$. On the other hand $c'$ will have to be proportional to $v$ twice. This contradiction implies that $c'$ can not coincide with $c$ for a reparameterization of $(\cos t, \sin 2t)$.