Confusion on homomorphism map

Question

For any group $G$, consider the map $f: G \to \text{Aut($G$)}$ with $g \mapsto f_g (h) = ghg^{-1}$. . This map represents a homomorphism from $G$ into itself, but that it satisfies the requirements are not immediately clear to me.

For example, it must be the case that $f$ maps the identity in $G$ to the identity in $\text{Aut($G$)}$. The identity in $\text{Aut($G$)}$ is the trivial map $f_e (h) = ehe^{-1} = h$. Since identities are unique, this must be the unique identity. However, the kernel of $f$ is the center of $G$, since if $g$ commutes with everything in the group, $ghg^{-1} = h$ for all $h \in G$.

I'm having trouble grasping this. The identity map in $\text{Aut($G$)}$ should not depend on the particular element of $G$, but by selecting $g$ that commute, it seems that we can induce an identity map.

Answer 1

$f$ maps $e$ to the identity automorphism, but it also maps any other element of the center to the identity.

For instance, the Klein four group is abelian, so $f$ would map everything to the identity (not just $e$, as it must to be a homomorphism).

Answer 2

Firstly, I think that a writing like "$g \mapsto f_g (h) = ghg^{-1}$" is misleading: $f$ takes an element of $G$ to an automorphism of $G$, not to another element of $G$. So, a better way is: $g \mapsto (h \mapsto f_g (h) := ghg^{-1})$. (Or, even better, use stacked maps.)

Said this, it's a fact that:

$$g \in Z(G) \Rightarrow f_g(h)=ghg^{-1}=hgg^{-1}=h, \color{red}{\forall h \in G} \Rightarrow f_g=\iota_G$$

IMO, you should restart from this to see what doesn't still convince you (if any).