How many field automorphisms does $\mathbf{C}$ have? If you assume the axiom of choice, there are tons of them -- $2^{2^{\aleph\_0}}$, I believe. And what if you don't -- how essential is the axiom of choice to constructing "wild" automorphisms of $\mathbf{C}$? Specifically, if you assume that $\mathsf{ZF}$ admits a model, does that imply that $\mathsf{ZF}$ admits a model where $\mathbf{C}$ has no wild automorphisms: $\operatorname{Aut}(\mathbf{C})=\mathbf{Z}/2\mathbf{Z}$?


I suppose if that's true, then the next logical question is to construct models of $\mathsf{ZF}$ where $\operatorname{Aut}(\mathbf{C})$ has cardinality strictly between $2$ and $2^{2^{\aleph\_0}}$--pretty disturbing if you ask me. Which finite groups can you hit?



Does $\operatorname{Con}\sf(ZF)$ imply $\operatorname{Con}\sf(ZF + \operatorname{Aut}{\bf C = Z/\mathrm 2Z})$?
The use of inaccessible cardinals is not necessary here, the Baire property works just as well as Lebesgue measure. Shelah (*Can you take Solovay's inaccessible away*, Isr. J. Math. 48, 1984, 1-47) shows that $\mathsf{ZF}$ + $\mathsf{DC}$ + "every subset of $\mathbf{R}$ has the [Baire property](http://en.wikipedia.org/wiki/Property_of_Baire)" is relatively consistent with $\mathsf{ZF}$. (This is also the paper where Shelah also shows that the inaccessible cardinal is necessary for Solovay's result.)


The connection is an old theorem of Banach and Pettis which says that any Baire measurable homomorphism between Polish groups is automatically continuous. This result is provable in $\mathsf{ZF}$ + $\mathsf{DC}$. Since $\mathbf{C}$ is a Polish group under addition, it follows that every additive endomorphism of $\mathbf{C}$ is continuous in Shelah's model. Since the continuous additive endomorphisms of $\mathbf{C}$ are precisely the $\mathbf{R}$-vector space endomorphisms, it follows that the only field automorphisms of $\mathbf{C}$ in Shelah's model are the identity and conjugation.




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As pointed out by Pete Clark in the comments, the Artin-Schreier Theorem goes through using only the Boolean Prime Ideal Theorem ($\mathsf{PIT}$), which is significantly weaker than full $\mathsf{AC}$. This shows that $\mathsf{AC}$ is not completely necessary to show that there is a unique conjugacy class of elements of order $2$ in $\mathrm{Aut}(\mathbf{C})$ and that these correspond precisely to the finite subgroups of $\mathrm{Aut}(\mathbf{C})$.


Looking at Pete Clark's [Field Theory Notes](http://alpha.math.uga.edu/%7Epete/FieldTheory.pdf), specifically at Steps 4 and 5 of his proof of the Grand Artin-Schreier Theorem on pages 62-63, I think that it is a theorem of $\mathsf{ZF}$ that the only possible order for a nontrivial finite subgroup of $\mathrm{Aut}(\mathbf{C})$ is $2$.


