Isometry group $\simeq \mathbb{Z}/2\mathbb{Z}$

Question

Is it possible to find a norm on $\mathbb{R}^2$ such that the linear isometry group consists only in $\pm \operatorname{id}$ ?

I think this is possible, but I didn't manage to construct such a unit ball.

Answer 1

Consider the norm with unit ball $$B=\{\,(x,y)\in\Bbb R^2\mid x^2+y^2\le1\land |x|\le 1-\epsilon\land |y|\le 1-\epsilon\land |3x+4y|\le 5-\epsilon\,\}. $$

This is a disk with some parts "chopped off". As it is pointsymmetric and convex, we can define a norm accordingly. Any linear map that maps $B$ to itself, must respect the circular arcs, hence can only be a rotation with or without a reflection (i.e., an orthogonal map). But it must also respect the "irregular" pattern of the straight line segments, which rules out everything but $\pm\operatorname{id}$.