= Solution
A <group automorphism> is a bijective <group homomorphism> from $G$ to itself. The multiplication in $\operatorname{Aut}(G)$ is composition: $(\alpha\beta)(g)=\alpha(\beta(g))$. Composition is associative, the identity map is the identity element, and each inverse <bijection> is again a <homomorphism>, so these maps form the <automorphism group>.
For conjugation by $h$, multiplication is preserved because
$$
h(g_1g_2)h^{-1}=(hg_1h^{-1})(hg_2h^{-1}).
$$
Conjugation by $h^{-1}$ is its inverse, so $\psi(h)$ is an <group automorphism>. Moreover
$$
\psi(h_1h_2)(g)=h_1h_2g h_2^{-1}h_1^{-1}
=\psi(h_1)(\psi(h_2)(g)).
$$
Hence \b[$\psi:G\to\operatorname{Aut}(G)$ is a <homomorphism>], with image the <inner automorphisms>. Its <group homomorphism kernel> is the <group center> of $G$, since $\psi(h)$ is the identity precisely when $hg=gh$ for every $g$.
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