A transfer map that destroys candidacy breaks exactly the exponential

Let $u$ be a candidate ($u \neq 0$, with $|u|$ and $e^{u}$ algebraic) and let $\Phi : \mathbb{C} \to \mathbb{C}$ be a ring homomorphism that fixes every algebraic number and satisfies $\Phi(\bar u) = \overline{\Phi(u)}$. If $\Phi(u)$ is not a candidate, then:

• $|\Phi(u)| = |u|$;
• $\Phi(e^{u}) \neq e^{\Phi(u)}$;
• $\Phi$ does not map $\mathbb{R}$ into $\mathbb{R}$;
• $\Phi$ does not commute with complex conjugation on all of $\mathbb{C}$;
• $\Phi$ is not continuous;
• $\Phi$ is neither the identity nor complex conjugation.

Proof sketch. $u\bar u = |u|^{2}$ is algebraic, so $\Phi$ fixes it and $\Phi(u)\overline{\Phi(u)} = |u|^{2}$, which gives the modulus. A ring endomorphism of $\mathbb{C}$ that maps $\mathbb{R}$ into $\mathbb{R}$ restricts to a ring endomorphism of $\mathbb{R}$, which is the identity; so $\Phi$ is determined by $\Phi(i) = \pm i$ and is the identity or conjugation. Commuting with conjugation everywhere forces $\Phi(\mathbb{R}) \subseteq \mathbb{R}$, and continuity gives the same conclusion, so each of these conditions is equivalent to $\Phi \in \{\mathrm{id}, \mathrm{conj}\}$. Both of those send candidates to candidates. Finally, if $\Phi(e^{u}) = e^{\Phi(u)}$, then $\Phi(u) \neq 0$, $|\Phi(u)| = |u|$ is algebraic and $e^{\Phi(u)} = \Phi(e^{u})$ is algebraic, so $\Phi(u)$ would be a candidate.

What it is for. Diaz.candidate_indistinguishable gives, for a hypothetical counterexample $u$, a map $\Phi$ of this kind with $\Phi(u)$ transcendental, hence not a candidate. This node says what such a map must fail to do. The modulus survives the transfer exactly; the only hypothesis that breaks is the exponential one, and the map is necessarily wild.

Not claimed. Nothing about Diaz's conjecture itself.