Is there a smooth distance-persaving bijection between two simple curves with same length? If $M,N \subset \mathbb R^n$ are two simple curves with same length $L[M]=L[N]<\infty$, is there a smooth bijection $F: M \to N$ so that for every parametrized curve $\gamma$ in $M$ $$L[\gamma]=L[F\circ \gamma]$$ holds?

$\newcommand{\Reals}{\mathbf{R}}$Just so this has an answer: Conceptually, $M$ and $N$ are isometric to a circle of fixed circumference, hence isometric to each other, and an isometry $F:M \to N$ has the desired property.
In more detail, let $\ell = L[M] = L[N]$ be the common lengths of $M$ and $N$. Pick a point $m$ in $M$ arbitrarily and an arc length parametrization $\mu:[0, \ell] \to M$ such that $\mu(0) = \mu(\ell) = m$. Extending $\mu$ by periodicity defines an isometry $\mu:\Reals/\ell\mathbf{Z} \to M$.
Similarly, if $n$ is an arbitrary point of $N$ and $\nu:[0, \ell] \to N$ is an arc length parametrization with $\nu(0) = \nu(\ell) = n$, then the periodic extension of $\nu$ defines an isometry $\nu:\Reals/\ell\mathbf{Z} \to N$.
The mapping $F = \nu \circ \mu^{-1}:M \to N$, defined by $F\bigl(\mu(t)\bigr) = \nu(t)$ for $t$ real, is therefore a composition of isometries, hence an isometry.
If $\gamma:I \to M$ is a curve, then $\|(F \circ \gamma)'\| = \|\gamma'\|$ because $F$ is an isometry, so $L[F \circ \gamma] = L[\gamma]$.
