Normal subgroup comparable with all normal subgroups implies characteristic in finite group

Statement
Suppose $$G$$ is a finite group and $$H$$ is a normal subgroup of $$G$$ such that for any normal subgroup K of $$G$$, either $$H \le K$$ or $$K \le H$$. Then, $$H$$ is a characteristic subgroup of $$G$$.

Applications

 * Comparable with all normal subgroups implies characteristic in finite nilpotent group

Breakdown for infinite groups

 * Normal subgroup comparable with all normal subgroups not implies characteristic

Proof
Given: A finite group $$G$$, a normal subgroup $$H$$ such that for any normal subgroup $$K$$ of $$G$$, either $$H \le K$$ or $$K \le H$$.

To prove: For any automorphism $$\sigma$$ of $$G$$, $$\sigma(H) = H$$.

Proof: We observe that:


 * Since $$G$$ is finite, $$H$$ is finite.
 * Since $$H$$ is normal and $$\sigma$$ is an automorphism, $$\sigma(H)$$ is also normal.
 * Thus, either $$H \le \sigma(H)$$ or $$\sigma(H) \le H$$.
 * Since $$\sigma(H)$$ is isomorphic to $$H$$, $$H$$ and $$\sigma(H)$$ have the same cardinality. Combining this with the previous fact yields that $$H = \sigma(H)$$.