Let $G$ be a group. For every $x\in G$ we define a mapping $$\phi_x:G\rightarrow G,\quad y\mapsto x y x^{-1},\quad y\in G,$$ called conjugation by $x$ It is easy to show the conjugation map is in fact, a group automorphism.

An automorphism of $G$ that corresponds to conjugation by some $x\in G$ is called inner. An automorphism that isn't inner is called an outer automorphism.

The composition operation gives the set of all automorphisms of $G$ the structure of a group, $\operatorname{Aut}(G)$ The inner automorphisms also form a group, $\operatorname{Inn}(G)$ which is a normal subgroup of $\operatorname{Aut}(G)$ Indeed, if $\phi_x,\; x\in G$ is an inner automorphism and $\pi:G\rightarrow G$ an arbitrary automorphism, then $$\pi\circ \phi_x \circ\pi^{-1} = \phi_{\pi(x)}.$$ Let us also note that the mapping $$x\mapsto \phi_x,\quad x\in G$$ is a surjective group homomorphism with kernel $\operatorname{Z}(G)$ the centre subgroup. Consequently, $\operatorname{Inn}(G)$ is naturally isomorphic to the quotient of $G/\operatorname{Z}(G)$

Note: the above definitions and assertions hold, mutatis mutandi, if we define the conjugation action of $x\in G$ on $B$ to be the right action $$ y\mapsto x^{-1} y x,\quad y\in G,$$ rather than the left action given above.