When are two inner automorphisms equal?

Question

Let $(G, \cdot) $ be a finite group. For $g\in G$, we define $i_g : G \to G, i_g=gxg^{-1}$. I would like to know when do we have that $i_a=i_b$ for $a \ne b$.
I managed to observe that if $a$ and $b$ are central then $i_a=i_b=i_e$.
I also saw that for some $a\in G$ we have that $i_a=i_{a^{-1}} \iff a^2 \in Z(G) $, but I couldn't make much progress on the general statement.

Answer 1

We have $axa^{-1}=bxb^{-1}\iff(b^{-1}a)x=x(b^{-1}a)$ for all $x\in G$, that is, $b^{-1}a$ commutes with every element of $G$.

Thus, we have that $b^{-1}a\in Z(G)$.

Answer 2

From a more abstract viewpoint than already noted, we have a group homomorphism $\phi: G \rightarrow \text{Inn}(G)$ given by $g \mapsto i_g$. For any group morphism $f$, $f(g)=f(h)$ iff $gh^{-1} \in \text{ker}(f)$. So we need to find the kernel of $\phi$. Well, $g \in \text{ker}(\phi)$ iff for all $x \in G$ $gxg^{-1}=x$ which is clearly (by multiplying both sides by $g$ ) equivalent to saying $g \in Z(G)$.