$\mathrm{Hom}_{\Bbb Z}(\Bbb C^*, \Bbb C^*)=\Bbb C^*$? I am aware of that $\mathrm{Hom}_{\Bbb Z}(X, X)=X$ if $X$ is either $\Bbb Z$ or $\Bbb Q$. Is this identity valid for $X=(\Bbb C^*, \times)$? A wider question may arise: what kind of abelian group $X$ that the identity is valid?
 A: The group $\mathbb{C}^*$ is isomorphic to $\mathbb{R}\times\mathbb{R}/\mathbb{Z}$, via
$$
z\mapsto\left(\log\lvert z\rvert,\frac{\arg{z}}{2\pi}+\mathbb{Z}\right)
$$
where the argument function takes values in $[0,2\pi)$. The groups $\mathbb{R}$ and $\mathbb{R}/\mathbb{Z}$ are additive. We now need to compute
\begin{align}\def\Hom{\operatorname{Hom}_{\mathbb{Z}}}
&\Hom(\mathbb{R},\mathbb{R}) \\
&\Hom(\mathbb{R}/\mathbb{Z},\mathbb{R}) \\
&\Hom(\mathbb{R},\mathbb{R}/\mathbb{Z}) \\
&\Hom(\mathbb{R}/\mathbb{Z},\mathbb{R}/\mathbb{Z})
\end{align}
Note that, as abelian group, $\mathbb{R}\cong\mathbb{Q}^{(c)}$ (a direct sum of $c=2^{\aleph_0}$ copies of $\mathbb{Q}$). Also, it can be proved that
$$
\mathbb{R}/\mathbb{Z}\cong\mathbb{Q}^{(c)}\oplus
\Bigl(\bigoplus_{p}\mathbb{Z}(p^{\infty})\Bigr)
$$
where $p$ runs through the prime numbers and $\mathbb{Z}(p^{\infty})$ is the Prüfer $p$-group.
Just the fact that $\Hom(\mathbb{R},\mathbb{R})$ has cardinality at least $2^c$ shows, by cardinality reasons, that your conjectured isomorphism doesn't hold.
There are other groups $G$ so that $\Hom(G,G)\cong G$, for instance all cyclic groups.
