Can we reparametrise a closed curve such that its derivative looks like the original curve? 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 $t \in \mathbb R$
 A: 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 $(\cos t, \sin 2t)$.
