Automorphisms of projective spaces, and the Axiom of Choice It is known that upon not accepting the Axiom of Choice (AC), there exist models of ZF in which there are projective spaces (over a division ring) with a trivial automorphism group. (This is a truly remarkable result, if I may say.)
Now let $G$, more generally, be any given group. Do there exist models of ZF (without AC) in which there are projective spaces $\mathbb{P}$ over some division ring, for which the automorphism group of $\mathbb{P}$ is isomorphic to $G$ ?
 A: On abstract metamathematical grounds (having nothing to do with projective spaces), I claim that it is relatively consistent with ZFC and indeed with ZFC+Con(ZF) and much more that the answer to your question is negative. This doesn't mean the answer is negative, since it could also be consistent that the answer is positive, making the property independent of ZFC+Con(ZF). It could also be that a positive answer to your question is a consequence of a stronger theory, such as ZFC plus large cardinals.
The fact that this answer has nothing at all to do with projective spaces shows that perhaps the question that was asked isn't precisely the question that one might have wanted to ask.
In particular, the role of Con(ZF) is important for your question as it was asked. Your initial remark is not quite stated correctly, since what we need for the initial result is not to reject AC, but (presumably) to assume Con(ZF), in order to get the model of ZF as you mentioned.
But I shall argue that neither Con(ZF) nor indeed any iteration of consistency statements Con${}^n$(ZF) is sufficient in general to get your more general statement. The main reason is that it is consistent with these theories that some groups $G$ simply cannot appear in a model of ZF at all.
To see this, observe first that Con(ZF) and the stronger iterated statements are strictly weaker than the assertion that there is an $\omega$-standard model of ZF. One can see that it is weaker because all $\omega$-standard models fulfill the same consistency statements as the ambient model, and so if Con${}^n$(ZF) implied the existence of an $\omega$-standard model of ZF, then it would imply its own consistency, contrary to the incompleteness theorem.
So if ZF+Con${}^n$(ZF) is consistent, then there is a model of this theory that thinks there are no $\omega$-standard models of ZF. In such a set-theoretic world, there are many models of ZF, but no $\omega$-standard models of ZF. But in this case, no model of ZF could have a group that is isomorphic to the standard integers $\mathbb{Z}$, because this can happen only in an $\omega$-standard model.
In summary, if ZF+Con(ZF) is consistent, then there are models of this theory in which the answer to your question is no.
