It's well known that there's no satisfactory analog of inner product spaces over non-Archimedean fields. A nice property of Hilbert spaces is that the automorphism group is highly transitive on the unit sphere in the following sense: given any two sets of points $x_1, x_2, ..., x_n$ and $y_1, y_2, ..., y_n$ for which $\lVert x_i\rVert = \lVert y_j \lVert = 1$ and $\lVert x_i - x_j \lVert = \lVert y_i - y_j\lVert$ for all $i,j\in[1,n]$ there is a (linear isometric) automorphism $L$ that satisfies $Lx_i = y_i$ for all $i\in[1,n]$. (Call this 'rigidly $n$-transitivite' for each $n$.)
How close to this amount of transitivity can you get in a Banach space over a non-Archimedean field? For starters is there even a non-Archimedean Banach space (with dimension $>1$, where the norm satisfies the ultrametric inequality as is typically studied) with automorphisms that are transitive on the unit sphere? Given the answer to this question I would imagine that in infinite dimensional spaces transitivity by itself is not a very strong constraint and that with no constraint on the base field and a sufficiently large dimension it ought to be possible. So can such a space be rigidly $n$-transitive for any $n\geq 1$?
This came up again and I realized there are such spaces and in fact there's a very nice family of non-Archimedean Banach spaces with this property, namely "$c_0(I)$" over any non-Archimedean field $K$ ($K$ may need to have a discrete value group), where $c_0(I)$ is the closure under the uniform norm of the set of functions $I\rightarrow K$ with finite support. In some sense these are the non-Archimedean analogs of Hilbert spaces: There's a natural notion of orthogonality in non-Archimedean Banach spaces and every space that admits an orthonormal basis is isomorphic to some $c_0(I)$. This is discussed in chapter 5 (around page 171) of van Rooij's "Non-Archimedean Functional Analysis."