The latest question to be asked at the Group Pub Forum is a classic: can every group be realised as the automorphism group of a group? The answer is no, and the canonical answer is the infinite cyclic group. See this answer of mine for the details. In the back-and-forth mailings Arturo Magidin made the following point, which I would never have realised:
A infinite group of exponent two may not have an automorphism of order two. It depends on the Axiom of Choice.
That is, if we assume choice then your group is the direct sum of infinitely many copies of $C_2$ and we can simply choose a pair of $C_2$s in the direct sum decomposition and switch them. Indeed, the direct sum will have automorphism group containing $S_{\infty}$. My question is therefore the following.
Question: Do not assume choice. What might the automorphism group of an infinite group of exponent two look like?
If it cannot be infinite cyclic then the canonical counter-example to the motivating question holds. But I don't really care about the motivating question, I am just interested on what is going on here...
(As an aside, I am not that comfortable with axioms - is "not assuming choice" the same as "assuming choice does not hold"? Is there some middle ground which sits before I decide to use choice or not? What happens here?...Basically, in your answer make it clear what is going on with your axioms!)
So to your question, it's hard to tell. Not assuming choice means that it might be true, in which case you already know the theory; or it might be false but its failure does not affect this; and so on.
It is, however, consistent that there exists an infinite abelian group of exponent $2$, or simply put, a vector space over $\Bbb F_2$, which does not admit any automorphism. It is consistent for such example to have the property that every proper subgroup is finite.
Note that this means that the group is not the direct sum of copies of $C_2$. However it does embed every such finite sum. If $G=\bigoplus_{i\in I}C_2$ then we can easily come up with an automorphism of $G$ which has order $2$. Pick some $i\in I$ and consider the automorphism switching the $i$-th copy of $C_2$, but keeping the rest in place.
The example above can be generalized, and we can have such group without automorphisms that every proper subgroup is the direct sum of copies of $C_2$ (and more specifically, the index set is well-orderable), but the group itself - again - is not.
You can see the details of the construction in Axiom of choice and automorphisms of vector spaces, which was later generalized to my masters thesis (which can be found through my user profile) to the generalized case.