= Solution
For fixed $g\in G$, conjugation $g_*:x\mapsto gxg^{-1}$ preserves multiplication:
$$
g_*(xy)=gxyg^{-1}=(gxg^{-1})(gyg^{-1}),
$$
and its inverse is $(g^{-1})_*$. It is therefore a <group automorphism>. Furthermore,
$$
(gh)_*(x)=ghx(gh)^{-1}=g_*(h_*(x)),
$$
so $\theta:g\mapsto g_*$ is a homomorphism $G\to\operatorname{Aut}(G)$. Its kernel consists of elements commuting with every $x$, hence
$$
\boxed{\ker\theta=Z(G),\qquad \operatorname{im}\theta=\operatorname{Inn}(G).}
$$
For $\alpha\in\operatorname{Aut}(G)$,
$$
\alpha g_*\alpha^{-1}=(\alpha(g))_*,
$$
so the <inner automorphism group> is normal in the <automorphism group>.
If $G=\langle x\rangle$ is cyclic, every automorphism has the form $x\mapsto x^a$. Two such maps commute because $ab=ba$, so $\operatorname{Aut}(G)$ is abelian. The conclusion fails for general abelian groups: for $G=C_2\times C_2$, its three nonidentity elements may be permuted arbitrarily, giving
$$
\boxed{\operatorname{Aut}(C_2\times C_2)\cong S_3,}
$$
which is nonabelian.
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