Let $\|\cdot\|$ be an arbitrary norm on $\mathbb R^n$, and suppose there exists an isometry $T:\mathbb R^n\to\mathbb R^n$ that is not simply a composition of reflections along some of the $n$ axes. That is, $T$ is linear, preserves $\|\cdot\|$, and there exists some standard basis element $e_i$ such that $Te_i\neq\pm e_i.$ Suppose further that $\|e_i\|=1$ for each $1\leqslant i\leqslant n,$ where $\{e_i\}$.
Conjecture: Under these assumptions $\|\cdot\|$ must be the standard Euclidean norm.
Is this true? My intuition says yes, but I have no idea where to begin proving it.
(This question is a follow-up to this one.)
A norm is fully characterized by its unit ball which must be compact convex and symmetric about the origin. An isometry must preserve the unit ball. Now, if you take the plane for instance and the unit ball to be a regular 2n polygon, it has some finite group of rotations as symmetries. These rotations may be products of reflections, but I don't think all of them are generated by reflections about the x and y axis.