By an absolute value over $\mathbb{C}$, I mean a function $\mathbb{C}\to \mathbb{R}_{\ge 0}$ such that:
$\bullet$ $|x| = 0\Longleftrightarrow x=0$;
$\bullet$ for all $x,y\in\mathbb{C}$ we have $|xy| = |x||y|$;
$\bullet$ for all $x,y\in\mathbb{C}$ we have $|x+y|\le |x|+|y|$.
An archimedean absolute value is one that does not satisfy the stronger inequality "$|x+y|\le \max\{|x|,|y|\}$ for all $x,y\in\mathbb{C}$". A complete absolute value is one inducing a complete metric.
Now, the two following assertions are equivalent:
(i) every archimedean absolute value over $\mathbb{C}$ is complete;
(ii) for every archimedean absolute value $|\cdot|'$ over $\mathbb{C}$, there exists an automorphism $\sigma$ of $\mathbb{C}$ and $c\in (0,1]$ such that $|x|' = |\sigma(x)|^c$ for all $x\in\mathbb{C}$, where $|\cdot|$ is the usual absolute value of $\mathbb{C}$.
((ii) $\Rightarrow$ (i) is obvious. (i) $\Rightarrow$ (ii) follows from one of the two Ostrowski's theorems.)
So, I was wondering if (i) holds, that is, every archimedean absolute value over $\mathbb{C}$ being complete? Thanks for your help.
The question is then whether you can get a $\sigma$ which is not the identity or the complex conjugation.
If you have the axiom of choice you can construct wild automorphisms, e.g. get a transcendence basis of $\mathbb{C}$ over $\mathbb{Q}$, permute it in such a way that one of its elements doesn't go to itself or its conjugate and extend the permutation to an automorphism.
If you don't have it, it is possible to have models of set theory where the only automorphisms of $\mathbb{C}$ are the identity and the complex conjugation.