Is it true that if I have $\alpha \in \mathbb{C}$, $q,w \in \mathbb{Z}$ such that for every automorphism $\sigma$ of $\mathbb{C}$, $$|\sigma(\alpha)|=q^{w/2}$$ then $\alpha$ must in fact be algebraic?


Note this is false without assuming some kind of axiom of choice, as else the identity and complex conjugation are the only automorphisms guaranteed to exist.



Norm fixed under complex automorphisms implies algebraic
Assuming ZFC, as an abstract field $\mathbb C$ is an algebraic closure of a pure transcendental extension of $\mathbb Q$ on $2^{\aleph\_0}$ generators. The automorphism group, of order $2^{2^{\aleph\_0}}$, is transitive on transcendental elements. So the answer is yes.


