Let $\mathbb{K}$ be the real numbers or the complex numbers, let $S$ be a set with at least two elements,
let $\hspace{.04 in}f : [0,\hspace{-0.04 in}\scriptsize+\normalsize \infty) \to [0,\hspace{-0.04 in}\scriptsize+\normalsize \infty)$ be an injection such that $\hspace{.04 in}f(1) = 1$ ,
let $V$ be the set of functions $v : S\to \mathbb{K}$ such that $\;\;\; \displaystyle\sum_{s\in S}$ $\hspace{.04 in}f(|v(s)|) \; < \; \scriptsize+\normalsize \infty \;\;\;$,
assume $V$ is a vector subspace of $\mathbb{K}^S$, and suppose that
$||\hspace{-0.05 in}\cdot\hspace{-0.05 in}|| : V\to [0,\hspace{-0.04 in}\scriptsize+\normalsize \infty)$ is a norm such that for all vectors $v$ in $V\hspace{-0.03 in}$, $\;\;\; \hspace{.04 in}f(||v||) \; =$ $\;\displaystyle\sum_{s\in S}$ $\hspace{.04 in}f(|v(s)|)$
and for that norm, there is an isometric automorphism which is not a
generalized permutation operator whose non-zero entries all have absolute-value $1$.
Does it follow that for all elements $x$ of $[0,\hspace{-0.04 in}\scriptsize+\normalsize \infty)$, $\hspace{.04 in}f(x) = x^2$ ?
(I came up with that question during linear algebra class when the professor mentioned that
the $p$-norm analogue of the Pythagorean theorem holds for vectors with disjoint support.)