Let $F$ be a meromorphic function on $\mathbb{C}$, and assume that the first-order theory of $(\mathbb{C},F)$ defines $\mathbb{Z}$, which means that there exists a formula $\varphi(z)$ (in the language of fields, so we cannot use complex conjugation, but parameters from $\mathbb{C}$ are allowed) such that
$\{z \in \mathbb{C}\mid \varphi(z)\}=\mathbb{Z}$.
(The last statement is not strictly first-order, but it can be made so: more precisely we can express that $\varphi$ defines a cyclic group that happens to be $\mathbb{Z}$ in the model $(\mathbb{C},F)$).
Now I propose the following statement:
$\mathbb{R}$ is not definable in $(\mathbb{C},F)$;
There are $2^{2^{\aleph_0}}$ automorphisms $\sigma$ of $\mathbb{C}$ such that $$\forall z \in \mathbb{C} (\sigma\circ F(z)=F \circ \sigma (z)).$$
Is the above statement plausible? Would it be of any possible interest?