Let $n \ge 2$ be some positive integer. Given a norm $p : \mathbb{R}^n \to \mathbb{R}$, one can inquire about the structure and properties of its isometry group, i.e. the group of all bijections $F:\mathbb{R}^{n}\to\mathbb{R}^{n}$ such that $p\left(v\right)=p\left(F\left(v\right)\right)$ for all $v \in \mathbb{R}^n$. By the Mazur-Ulam theorem, the isometries of $p$ are affine transformations, so the subgroup $Iso\_{0}\left(p\right)$ of all isometries which fix the origin, consists of linear transformations, and is therefore a closed subgroup of $GL\_n (\mathbb{R})$. Essentially, $Iso\_{0}\left(p\right)$ is what we get after modding out the obvious distance-preserving functions (the translations), which have no relation to $p$ whatsoever.


For some norms, $Iso\_{0}\left(p\right)$ is a very small group. For instance, if $p$ is either the $\ell^\infty$ norm or the $\ell^1$ norm, then the group of $Iso\_{0}\left(p\right)$ is finite, and this seems to be the case for all $\ell^q$ norms other than $q=2$. On the other hand, for the euclidean norm ($p = \ell^2$), $Iso\_{0}\left(p\right)$ is quite a rich group: it is a Lie group of positive dimension (which increases significantly with $n$) - the orthogonal group $O(n)$. 


Obviously, a significant difference between this norm and other possible norms is that it arises from an inner product on $\mathbb{R}^n$. Since there is only one inner product structure on $\mathbb{R}^n$ (up to conjugacy) this essentially gives us just one example of a norm with a rich group of linear isometries. So my questions are:


1. Are there any other norms (not induced by an inner product) on $\mathbb{R}^n$ which have a rich group of linear isometries? Here let me define "rich" as having positive dimension as a Lie subgroup of $GL\_n (\mathbb{R})$ (every closed linear group is canonically a submanifold of $GL\_n (\mathbb{R})$ compatible with the group structure).
2. If not, why is the property of being induced from an inner product the "right" condition for having many symmetries? That is, from a geometric point of view, why does this property make the norm especially "symmetric" or "smooth"?



Which norms have rich isometry groups?
As an answer to (1), consider any compact subgroup $G\subset O(n)$ and look at the $G$-invariant functions on $\mathbb{R}^n$ that have homogeneity $1$. As long as $G$ does not act transitively on the space of lines in $\mathbb{R}^n$, that set of functions will be properly larger than the $O(n)$-invariant functions of homogeneity $1$, and any one of them that is sufficiently near the standard norm will be a norm. 


As a specific example, consider the irreducible action of $SO(3)$ on the $5$-dimensional vector space $S$ of traceless symmetric $3$-by-$3$ matrices. There are two $SO(3)$-invariant polynomials on $S$: $p\_2(m) = trace(m^2)$ and $p\_3(m) = det(m)$. (They generate the ring of $SO(3)$-invariant polynomials on $S$.) Now let 
$$
|m| = \bigl(p\_2(m)^3 + \epsilon\ p\_3(m)^2\bigr)^{1/6}.
$$
for some small $\epsilon$. When $\epsilon>0$ is sufficiently small, this will be a strictly convex norm on $S$, and its symmetry group will be $SO(3)$, which acts irreducibly on $S$. 


