Let $G$ be a group. Then $G$ acts on itself by conjugation, which corresponds to a homomorphism $K\colon G\to\operatorname{Aut}(G)$. Show that the kernel of $K$ is $Z(G)$.
$K: G\to \operatorname{Aut}(G)$
$G\times G\to G$, $(g,x)\mapsto xgx^{−1}$.
$\ker(K)=Z(G)$? $\ker(K)=\{x\in G\mid K(x)=I_G\}$. How can I continue?