Consider the $p$-adic field $\mathbb{Q}_p$ and $n\geq 1$. The absolute value is $|x|=p^{-v_p(x)}$ for $x\in\mathbb{Q}_p$, where $v_p$ is the $p$-adic valuation.
Endow $V=\mathbb{Q}_p^n$ with the sup-norm: For $v\in V$, $\|v\|=\max_{i=1}^{n} |v_i|$.
Let $A\in M_{n\times n}(\mathbb{Z}_p)$.
Does the condition $v_p(\det A)=0$ imply that $\|Av\|=\|v\|$ for all $v\in V$? Is $v_p(\det A)=0$ also a necessary condition?
This is trivially true in the case $n=1$, and maybe I'm grossly generalizing, but I think this has a change of being true.
Yes, and more: The subgroup of those elements of $GL_n(K)$ which leave the sup-norm invariant is exactly $GL_n(\mathcal{O}_K)$, for any local field $K$ with its ring of integers $\mathcal{O}_K$. Cf. reuns' answer to Natural Extensions of the $p$-Adic Norm to Higher Dimensions . (And for any commutative ring $R$, $GL_n(R) =\{g\in M_n(R): \mathrm{det}(g) \in R^\times\}$. Cf. also Compact subgroups of $p$-adic fields and the groups $GL_n$ over them .)