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 ZF admits a model, does that imply that ZF admits a model where $\mathbf{C}$ has no wild automorphisms: $\mathop{Aut}\mathbf{C}=\mathbf{Z}/2\mathbf{Z}$?
I suppose if that's true, then the next logical question is to construct models of ZF where $\mathop{Aut}\mathbf{C}$ has cardinality strictly between 2 and $2^{2^{\aleph_0}}$--pretty disturbing if you ask me. Which finite groups can you hit?