Continuous automorphism groups of normed vector spaces? Consider the metric space on, say, ℝ2 induced by the various $L^p$ norms, and the group of isometries from that space into itself that preserve the origin. When $p=2$ I get the continuous group of rotations, but when $p\in\{1,3,4,5,...\infty\}$ it looks like I just get $D_8$, the symmetry group of the square. Question: what's going on here? Why is 2 so special? Are there other natural norms on ℝ2 (or on ℝn) besides the euclidean one that give interesting isometry groups?
 A: The following answer gives a partial description of the isometry groups of finite-dimensional normed spaces.
I assume that an isometry is a bijection preserving the distance function. By the Mazur-Ulam theorem it then follows that an isometry is a linear transformation composed with a translation. Thus we may assume without loss of generality that an isometry fixes the origin, so the isometry group is a subgroup of $GL(n)$.
Then the isometry group of any (real) finite-dimensional normed space is conjugate in $GL(n)$ to a closed subgroup of $O(n)$ that contain $-id$. This is seen as follows.
Consider the John ellipsoid $E$ of the unit ball $B$ of some $n$-dimensional normed space. This is the ellipsoid of largest volume contained in $B$ and, crucially, it is the unique such ellipsoid.
After some choice of basis we may assume that $E$ is the Euclidean ball. An isometry maps $B$ onto $B$, so it must map the John ellipsoid to the John ellipsoid. It follows that the isometry group is a subgroup of $O(n)$ containing $-id$. This subgroup is clearly closed, hence compact.
The converse is surely false. The following is an attempt at constructing a norm from such a subgroup. Fix a Euclidean unit vector $v$. Then its $Gv$ is a compact set of Euclidean unit vectors, symmetric with respect to the origin. Its convex hull $Gv$ is still compact and symmetric, so gives a unit ball $B_0$ of some norm on the linear span of $Gv$. If this linear span is not all of $\mathbb{R}^n$, then the unit ball has to be made full-dimensional in a sufficiently rough way, so as not to add any more isometries.
However, as pointed out by Leonid Kovalev in the comments, there are closed subgroups of $O(n)$, such as $U(n)$, where this construction gives a norm with a strictly larger isometry group (in the case of $U(n)$, the Euclidean norm).
As pointed out by Bill Johnson in a comment to his answer, it was shown by Gordon and Loewy that any $finite$ subgroup of $O(n)$ that contains $-id$ is the isometry group of some norm on $\mathbb{R}^n$. It's still my guess that the only way you can get infinite isometry groups (in the finite-dimensional case) is by having Euclidean subspaces, and for the norm to be so symmetric that it shares all the symmetries of this subspace.
A: I think what groups can be the isometry group of a finite dimensional normed space are classified, maybe by Y. Gordon and/or D.R. Lewis. I don't have access to emath from home but will check the reference tomorrow if no one has answered by then.
BTW:   Banach-spaces would be a more appropriate tag IMO.
A: Consider the following norm on $\mathbb{R}^{2}$:                                    $||(x,y)||$ := $|x|+|y|$
if $xy\leq0$; $||(x,y)||$ := $|y|$
if $xy\geq0$ and $|y|$ $\geq3|x|$; 
$||(x,y)||$ := $|x|+\frac{2}{3}|y|$
if $xy$ $>0$ and $|y|$ $\leq3|x|$. Then the group of isometries
is { $\pm I\ $}.
