Normal upper-hook fully normalized implies characteristic

Statement
If $$K \le H \le G$$ are groups, with $$K$$ normal in $$G$$, and $$H$$ fully normalized in $$G$$, then $$K$$ is characteristic in $$H$$.

Other related facts

 * Characteristic upper-hook AEP implies characteristic
 * Fully invariant upper-hook EEP implies fully invariant

Applications

 * Left transiter of normal is characteristic can be derived from the statement of this page, along with the fact that every group is normal fully normalized in its holomorph.
 * Characteristically simple and normal fully normalized implies minimal normal

Proof
Given: $$K \le H \le G$$ are groups, with $$K$$ normal in $$G$$, and $$H$$ fully normalized in $$G$$.

To prove: $$K$$ is characteristic in $$H$$.

Proof: Let $$\sigma$$ be any automorphism of $$H$$. We need to show that $$\sigma(K) = K$$.

Since $$H$$ is fully normalized in $$G$$, there exists $$g \in G$$ such that $$\sigma$$ equals conjugation by $$g$$. Since $$K$$ is normal in $$G$$, conjugation by $$g$$ leaves $$K$$ invariant, so $$\sigma(K) = K$$, completing the proof.