I was recently trying to think of a simple example that demonstrates that the natural inclusion of an abelian group $G$ into 
$$\widehat{\widehat{G}}=\text{Hom}\\_{\mathsf{Ab}}(\text{Hom}\_{\mathsf{Ab}}(G,\mathbb{C}^\times),\mathbb{C}^\times)$$
is not necessarily an isomorphism. Note that I'm looking at all homomorphisms; no topology on the group is involved (or, if you prefer, they are all discrete).


Obviously, $G$ has to be infinite. However, the only example I could think of (much less actually prove worked) was $G=\mathbb{Z}$, in which case 
$\widehat{G}=\text{Hom}\\_{\mathsf{Ab}}(\mathbb{Z},\mathbb{C}^\times)\cong\mathbb{C}^\times$, so that
$$\widehat{\widehat{G}}\cong\text{Hom}\_{\mathsf{Ab}}(\mathbb{C}^\times,\mathbb{C}^\times)$$
which is uncountable due to the existence of uncountably many automorphisms of the field $\mathbb{C}$, and therefore not isomorphic to $\mathbb{Z}$. However, (my understanding is that) the existence of the anything more than the two obvious automorphisms of $\mathbb{C}$ requires the axiom of choice. So my questions are,


Does the claim that $\mathbb{Z}\not\cong\text{Hom}\_{\mathsf{Ab}}(\mathbb{C}^\times,\mathbb{C}^\times)$ require AC? 


and, if the answer to the above is yes,


Does the claim that there exists *some* abelian group $G$ such that $G\not\cong\widehat{\widehat{G}}$ require AC? Is it equivalent?


I initially [asked this](https://math.stackexchange.com/questions/24091) on math.SE but I suspect it's harder than I initially thought.



The natural inclusion of an infinite abelian group $G$ into $\widehat{\widehat{G}}$
Since $\mathbb{C}^\times \cong \mathbb{R}\\_{+}\times S^1$ by polar coordinates, it suffices to show that $\text{Hom}({\mathbb R}\\_+,{\mathbb R}\\_+)$ is uncountable. But for any real number $a$, $x\mapsto x^a$ gives an endomorphism of ${\mathbb R}\_+$. Explicitly, there are uncountably many group homomorphisms from ${\mathbb C}^\times$ to itself given by $re^{i\theta}\mapsto r^ae^{i\theta}$, for any real $a$.


