Inner Automorphism Maps Subgroup to Itself iff Normal

Theorem
Let $G$ be a group.

Let $x \in G$.

Let $\kappa_x$ be the inner automorphism of $x$ in $G$.

Let $N$ be a normal subgroup of $G$.

Then:
 * $\kappa_x \sqbrk N = N$

Proof
By definition, $\kappa_x: G \to G$ is a mapping defined as:
 * $\forall g \in G: \map {\kappa_x} g = x g x^{-1}$

Let $n \in N$.

Then: