I'll prove the result in the following equivalent form:
THEOREM 0 Let $\ \alpha\ \beta\subseteq\mathbb R^2\ $ be two closed convex curves of the same (positive) length. Then either $\ \alpha\ \beta\ $ are isometric as subsets of the Euclidean plane $\ \mathbb R^2\ $ or there does not exist a metric function (i.e. Lip$_1)\ $ of any of the two curves onto the other one.
To this end, let's apply two classic results about curves:
THEOREM 1 The integral of the curve curvature over an arbitrary closed convex curve is always $\ 2\!\cdot\!\pi.$
This theorem holds for all closed convex curves with the following understanding: the curves corner points serve as measure atoms, where the integral over a corner is the angle between the extreme tangents at the given corner (in the case of a smooth point this angle is $\ 0$ (zero). (There is an extra to it about the orientation, but never mind).
Now, another classical theorem (a bit harder?): a continuous bijection $\ f\ $ of a plane curve onto another is an isometry $\ \Leftarrow:\Rightarrow\ $ the curvature at $\ x $ and $\ f(x)\ $ is the same at every point of the domain of $\ f$.
Now, the proof proper of THEOREM 0:
If the curves $\ \alpha\ \beta\ $ are isometric then there is nothing to prove.
Otherwise, consider an arbitrary curve-length preserving bijection $\ f :\alpha\rightarrow\beta\ $ hence $\ f\ $ is not an isometry (of subsets of $\ \mathbb R^2).\ $ Thus there exist points $\ x\ y\in\alpha\ $ such that
$$ \kappa_{\alpha}(x)<\kappa_{\beta}(f(x))\qquad\mbox{and}\qquad \kappa_{\alpha}(y)>\kappa_{\beta}(f(y)) $$
(where $\ \kappa$'s are the respective curvatures). These inequalities may happen for the corner points too.
We see that $\ f^{-1}\ $ is not locally metric around point $\ f(x),\ $ and $\ f\ $ is not locally metric at point $\ y.\ $ This proves THEOREM 0.