TITLE: Equivalence classes of norms on $R^n$ under symmetries
QUESTION [6 upvotes]: Let $G \leq {\bf GL}_n$ be a symmetry group on $\mathbb{R}^n$. For simplicity, we can consider the case $G = {\bf GL}_n$.
Define two norms $\|\cdot\|_1$ and $\| \cdot\|_2$ to be equivalent under $G$ if there exists $A \in G$ such that
$ \ \forall \ x \ \| A x \|_1 = \| x\|_2$
It is trivial to show that this defines equivalence classes on the set of all norms on $\mathbb{R}^n$.
Now, for $\mathbb{R}^1$, there is only a single class of norms but for $\mathbb{R}^2$ there are infinitely many classes, specifically all $p$ norms live in different equivalence classes (except for the fact that the $1$-norm and the $\infty$-norm are equivalent). For general $\mathbb{R}^n$ the case is even "worse".
Is there some way to classify these equivalence classes?

REPLY [6 votes]: When $G={\bf GL}_n$, the set of classes of equivalent norms over ${\mathbb R}^n$ is a compact Hausdorff space when equipped with the Banach-Mazur metric
$$d(N_1,N_2)=\log\delta(N_1,N_2)$$
where 
$$\delta(N_1,N_2)=\inf_{T\in{\bf GL}_n}\|T^{-1}\|_{N_2\rightarrow N_1}\cdot\|T\|_{N_1\rightarrow N_2}\,.$$
See this page.
Of course, the smaller the group $G$, the bigger the set of equivalence classes.