When matrix $A$ is linear isometry in $\|\cdot\|_{\infty}$ norm?

Question

What are necessary/sufficient conditions for matrix $A \in \mathbb{R}^{n\times n}$ to hold the following property? $$\|Av\|_{\infty} = \|v\|_{\infty}$$

Answer 1

It holds if and only if $A$ is an entrywise signed permutation matrix.

Since $\|Ae_j\|_\infty=\|e_j\|_\infty=1$, every $|a_{ij}|$ is bounded above by $1$ and each column of $A$ contains at least one entry whose value is $\pm1$.

On the other hand, as $\|v\|_\infty=\|Av\|_\infty$ for all $v$, we also have $\max_i\sum_j|a_{ij}|=\|A\|_\infty=1$. Therefore, each row of $A$ has at most one entry whose value is $\pm1$.

It follows that on each column or on each row of $A$, there is exactly one entry whose value is $\pm1$. However, as $\max_i\sum_j|a_{ij}|=\|A\|_\infty=1$, all other entries must be zero. Therefore $A$ is a permutation matrix carrying signs on its entries.