Pete's answer is informative. But there is a subtle
point that actually turns the answer somewhat upside down. It turns out that the answer is related to large cardinals! [Edit: François's answer shows how to avoid the inaccessible cardinal.]
Pete mentions the Kestelman article Automorphisms of the field of complex numbers, which explains
Every function which defines a non-trivial automorphism of
the complex numbers transforms every bounded set (in the
Argand plane) into a set of Lebesgue measure zero or else
into a non-measurable set.
By considering larger and larger bounded sets, this means
that the existence of a nontrivial automorphism implies the
existence of a nonmeasurable set. I believe that this part
of the Kestelman article does not use AC, although I
suppose that one must have Dependent Choices (DC) to have a
decent theory of Lebesgue measure.
Pete mentions that there are known to be models of ZF in
which every set of reals is measurable. These models,
however, as Gerald mentions in his comments, are
constructed from a ground model of ZFC having an
inaccessible cardinal (Solovay's model). Shelah has proved
that this large cardinal hypothesis cannot be omitted.
Thus, the consistency of ZF + DC + "Every set is Lebesgue
measurable" is equivalent to the theory ZFC + "there is an
inaccessible cardinal". One way to explain what this means
is that we should be exactly as confident in the
consistency of inaccessible cardinals as we are that there
is no analogue of the Vitali construction of a non-Lebesgue
measurable set not using AC.
Since the Kestelman result shows that the existence of a
nontrivial automorphism of C (in the presence of DC)
implies the existence of a nonmeasurable set, this
- Con(ZFC + "there is an inaccessible cardinal") implies
Con(ZF + DC + "there is no nontrivial automorphism of
This is the actual result that Pete's argument provides. The hypothesis here is strictly stronger than Con(ZF), if ZF is consistent. [Edit: François shows that by using the Baire property instead of measure, one avoids the need for inaccessible cardinals, so he has the optimal argument.]
Having DC in the conclusion seems what should be desired,
when considering functions on R and C, since even to know
that the epsilon–delta and convergent sequence
characterizations of continuity are equivalent uses DC.
I'm not sure what happens if one drops DC in the
conclusion. For example, it is known to be consistent with
ZF that the reals are a countable union of countable sets,
and this model does not have DC or even countable choice.
Perhaps this is a good candidate model?
Finally, the question about realizing other groups is