Condition for Group Endomorphism to Commute with All Inner Automorphisms

Theorem
Let $G$ be a group.

Let $\phi: G \to G$ be an endomorphism on $G$.

Let $\phi$ be such that:
 * $\forall a \in G: \kappa_a \circ \phi = \phi \circ \kappa_a$

where:
 * $\kappa_a$ denotes the inner automorphism of $G$ given by $a$
 * $\circ$ denotes composition of mappings.

Then:
 * $H = \set {x \in G: \map \phi {\map \phi x} = \map \phi x}$

is a normal subgroup of $G$.

Also, the quotient group $G / H$ is an abelian group.

Proof
We have for all $a, g \in G$:

Let $x \in H$.

Then by the above:

Hence $a x a^{-1} \in H$, and thus $a H a^{-1} \subseteq H$.

By definition of a normal subgroup, $H$ is a normal subgroup of $G$.

It remains to show that the quotient group $G / H$ is an abelian group.

We need to show that for any $a, b \in G$, we have:
 * $a H b H = b H a H$

that is,
 * $a b H = b a H$

By Cosets are Equal iff Product with Inverse in Subgroup, this is equivalent to:
 * $\paren {b a}^{-1} a b = a^{-1} b^{-1} a b \in H$

We have:

and thus $a^{-1} b^{-1} a b \in H$.

Hence the result.