What does it mean for an element of a group to be a function on the group?

Note: Each element $a \in G$ defines an automorphishm $$\phi_a : G \to G \,;\, \phi_a(x)=axa^{-1}$$

The mapping $$F : G \to Aut(G) \,;\, F(a)=\phi_a$$ is a group homomorphism. Therefore, by the first isomorphism Theorem, we get an isomorphism: $$F : G / \ker(F) \to Im(F) \leq Aut(G)$$ and it is straight forward to check $\ker(F)=\mathcal{Z}(G)$.

Now, via this "natural" mapping, $G/\mathcal{Z}(G)$ can be identified (is isomorphic) with a subgroup of $Aut(G)$, and we usually consider them to be "the same".

It would be better

which includes a normal subgroup isomorphic to $G/\mathcal{Z}(G)$

Given $g\in G$, you can consider the inner automorphism $\phi_g$ defined by $$ \phi_g(x)=gxg^{-1} $$ It is easy to see this is indeed an automorphism of $G$. Moreover $$ \phi_g\circ\phi_h=\phi_{gh} $$ so the map $\phi\colon G\to\operatorname{Aut}G$ defined by $g\mapsto \phi_g$ is a group homomorphism. Its kernel is $\mathcal{Z}(G)$, so the homomorphism theorem says we have an injective homomorphism $$ G/\mathcal{Z}(G)\to\operatorname{Aut}G $$ It remains to show the image is a normal subgroup.

You're right: it results from this identification. More precisely, any group acts on itself by inner automorphisms (one also says by conjugation).

There is a canonical homomorphism $\begin{aligned}[t]\varphi \colon G&\longrightarrow \operatorname{Aut}G\\g&\longmapsto(x\mapsto gxg^{-1})\end{aligned}$ The kernel of this homomorphism is, by definition, the centre of $G$, $Z(G)$, so that we obtain an embedding $\;G/Z(G)\hookrightarrow \operatorname{Aut}G$.

The image (the inner automorphisms) are a normal subgroup of $\operatorname{Aut}G$, because one easily checks that, if $f$ is an automorphism $G$, the conjugate of the inner automorphism induced by $g$ is the inner automorphism induced by $\varphi(g)$.