isometrygroup of hypercube

Question

Let $n \in \mathbb{N}$. I want to show, that an isometry $\alpha \in Isom([-1,1]^n)$ is already determined by it's action on the vertices. I really just want to use the invariance of distance. I don't want to use the fact, that $\alpha$ has to be linear.

Answer 1

Let $V$ be the vertex set. The map $(x,y)\mapsto d(x,y)$ where $d$ is the Euclidean distance is maximized for $x$ and $y$ in the cube when $x$ and $y$ are opposite vertices. So any isometry fixes the vertex set. Now it is reasonably straightforward that an isometry restricts to a Euclidean transformation on the vertex set. Finally each point in the cube is determined by the distances to the vertices, so the isometry must be a Euclidean transformation on the whole cube.