Here is a problem that has been bugging me for a while.

Let $\| \|$ be a norm over $\mathbb{R}^n$, let $C$ be a convex subset of $\mathbb{R}^n$ with non-empty interior, and let $f: (C,\|\|) \rightarrow (\mathbb{R}^n,\|\|)$ be a distance-preserving map.

Is it true that there exists an isometry $g$ of $(\mathbb{R}^n,\|\|)$, such that $f = g|_C$?

**A few notes:**

-I don't require an isometry to be surjective, though it follows from the bounded-compactness of $(\mathbb{R}^n,\|\|)$.

-There is flexibility regarding the caracteristics of $C$, which may (to serve my purpose) be: $C$ is an open ball or closed ball with non-empty interior. In that case, $f$ is a bijection from $C = B(x,r)$ onto $B(f(x),r)$.

-Mazur-Ulam's theorem states that every surjective isometry between two real normed vector-spaces is affine.

-Though it may not seem so, I think the easiest way to prove this is to show that $f$ perserves the middles of any two points of $C$. Unfortunately, this isn't an easy task, I do not think I would have been able to prove it in the case when $C = \mathbb{R}^n$ if I hadn't come across PFDs about Mazur-Ulam. I have failed to adapt the proof(s) of Mazur-Ulam theorem to this case. I don't understand the "fundamental reason" why this theorem works, making it difficult to use. Maybe it's just magic.

-The norm may not be strictly convex, therefore, given two distinct points, there may be infinitely many points whose distance to each of the original points is half the distance between the original points.

-There might be a solution if we assume that $f$ is differentiable on the interior of $C$, but once again I don't think it would be easy to prove that $f$ is differentiable without having established results which directly imply that it is affine.

Any idea how to proceed?