show/hide this revision's text 4 Improvement in light of comments.

The following answer attempts to characterize all gives a partial description of the isometry groups of finite-dimensional normed spacesas essentially the closed subgroups of $O(n)$. Perhaps someone could explain which these are, but I guess they are all "finite modulo Euclidean subspaces".

Then I think the isometry groups group of any (real) finite-dimensional normed spaces are exactly the conjugates space is conjugate in $GL(n)$ of the to a closed subgroups subgroup of $O(n)$ that contain $-id$. This is seen as follows.

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 necessarily closed subgroup of $O(n)$ containing $-id$.

I'm not so sure, but conversely, take any 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$G$ of $O(n)$ containing $-id$. . Fix such an orbit $Gv$ by taking $v$ to be a Euclidean unit vector . $v$. Then its $Gv$ is a compact set of Euclidean unit vectors. Then , symmetric with respect to the origin. Its convex hull of $Gv\cup -Gv$ 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 all of $\mathbb{R}^n$, then hopefully the only isometries are the elements of $G$. If the 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.
show/hide this revision's text 3 Added requirement that an isometry fixes the origin.

The following answer attempts to characterize all isometry groups of finite-dimensional normed spaces as essentially the closed subgroups of $O(n)$. Perhaps someone could explain which these are, but I guess they are all "finite modulo Euclidean subspaces".

I assume that an isometry is a bijection preserving the distance function. By the Mazur-Ulam theorem it then follows that all isometries are 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 I think the isometry groups of finite-dimensional normed spaces are exactly the conjugates in $GL(n)$ of the closed subgroups of $O(n)$ that contain $-id$.

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 necessarily closed subgroup of $O(n)$ containing $-id$.

I'm not so sure, but conversely, take any closed (hence compact) subgroup $G$ of $O(n)$ containing $-id$. Fix such an orbit $Gv$ by taking $v$ to be a Euclidean unit vector. Then $Gv$ is a compact set of Euclidean unit vectors. Then the convex hull of $Gv\cup -Gv$ is still compact, so gives a unit ball $B_0$ of some norm on the linear span of $Gv$. If this linear span is all of $\mathbb{R}^n$, then hopefully the only isometries are the elements of $G$. If the 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.
show/hide this revision's text 2 Added introductory sentence explaining the relation to the question.

The following answer attempts to characterize all isometry groups of finite-dimensional normed spaces as essentially the closed subgroups of $O(n)$. Perhaps someone could explain which these are, but I guess they are all "finite modulo Euclidean subspaces".

I assume that an isometry is a bijection preserving the distance function. By the Mazur-Ulam theorem it then follows that all isometries are linear.

Then I think the isometry groups of finite-dimensional normed spaces are exactly the conjugates in $GL(n)$ of the closed subgroups of $O(n)$ that contain $-id$.

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 necessarily closed subgroup of $O(n)$ containing $-id$.

I'm not so sure, but conversely, take any closed (hence compact) subgroup $G$ of $O(n)$ containing $-id$. Fix such an orbit $Gv$ by taking $v$ to be a Euclidean unit vector. Then $Gv$ is a compact set of Euclidean unit vectors. Then the convex hull of $Gv\cup -Gv$ is still compact, so gives a unit ball $B_0$ of some norm on the linear span of $Gv$. If this linear span is all of $\mathbb{R}^n$, then hopefully the only isometries are the elements of $G$. If the 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.
show/hide this revision's text 1