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So the problem is that your map $f$ is not surjective.

However, $f$ is locally surjective in the following sence:

If $\Omega$ is the interior of $C$, $x\in \Omega$ and $\varepsilon>0$ is such that $B_\varepsilon(x)\subset \Omega$ then the restriction $f|_{B_\varepsilon(x)}$ is a bijection from $B_\varepsilon(x)\to B_\varepsilon(f(x))$. This is true since the distance preserving map has to be volume preserving and $\mathrm{vol}\,B_\varepsilon(x)=\mathrm{vol}\,B_\varepsilon(f(x))$.

This local surjectivity can exchange the surjectivity in the proof of Jvaisala given here (the only proof of Mazur--Ulam's theorem I know). It gives the following: assume $B_{100{\cdot}|x-y|}(x)\subset \Omega$ then $$f\left(\frac{x+y}2\right)=\frac{f(x)+f(y)}{2},$$ i.e., the map $f$ is a restriction of affine map to $C$. Hence everything follows.

So the problem is that your map $f$ is not surjective.

However, $f$ is locally surjective in the following sence:

If $\Omega$ is the interior of $C$, $x\in \Omega$ and $\varepsilon>0$ is such that $B_\varepsilon(x)\subset \Omega$ then the restriction $f|_{B_\varepsilon(x)}$ is a bijection from $B_\varepsilon(x)\to B_\varepsilon(f(x))$. This is true since the distance preserving map has to be volume preserving and $\mathrm{vol}\,B_\varepsilon(x)=\mathrm{vol}\,B_\varepsilon(f(x))$.

This local surjectivity can exchange the surjectivity in the proof of Jväisälä given here (the only proof of Mazur–Ulam's theorem I know). It gives the following: assume $B_{100{\cdot}|x-y|}(x)\subset \Omega$ then $$f\left(\frac{x+y}2\right)=\frac{f(x)+f(y)}{2},$$ i.e., the map $f$ is a restriction of affine map to $C$. Hence everything follows.

Sorry, I did not read your question to the end and gave this answer;.

So the problem is that your map $f$ is not surjective.

However, $f$ is locally surjective in the following sence:

If $\Omega$ is the interior of $C$, $x\in \Omega$ and $\varepsilon>0$ is such that $B_\varepsilon(x)\subset \Omega$ then the restriction $f|_{B_\varepsilon(x)}$ is a bijection from $B_\varepsilon(x)\to B_\varepsilon(f(x))$. This is true since the distance preserving map has to be volume preserving and $\mathrm{vol}\,B_\varepsilon(x)=\mathrm{vol}\,B_\varepsilon(f(x))$.

This local surjectivity can exchange the surjectivity in the proof of Jvaisala given here (the only proof of Mazur--Ulam's theorem I know). It gives the following: assume $B_{100{\cdot}|x-y|}(x)\subset \Omega$ then $$f\left(\frac{x+y}2\right)=\frac{f(x)+f(y)}{2},$$ i.e., the map $f$ is a restriction of affine map to $C$. Hence everything follows.

In other words, you ask if any isometry of $(\mathbb R^n,\|{*}\|)$ is linear.

Yes this is true, I am sure it is written somewhere, but I do not know a reference, so let me prove it.

First note that if the unit ball is not polyhedron then you get enough directions with unique geodesics and everything is easy.

For metrics with polyhedral balls, you can proceed by induction on $n$. In this case the boundary of ball maps to the boundary of ball, for appropriate choice of point you can use it to show that some hyperplanes map to hyperplanes; this gives you the reduction of dimension.

Sorry, I did not read your question to the end and gave this answer;.