Subgroup maximal in its normalizer

Question

Let $G$ be a finite group, $H$ a subgroup of $G$, such that $H$ is a proper subgroup of $\operatorname{N}_G(H)$. Is there a sufficient condition for $H$ to be maximal in $\operatorname{N}_G(H)$?

What if $H$ is a $p$-Sylow subgroup of $G$, for some prime $p$? What if $H$ is furthermore of order $p$? What if $G$ is simple?

Answer 1

In the simple group ${\rm PSL}(2,11)$, a Sylow $3$-subgroup has order $3$, and its normalizer has order $12$, so it is not maximal in its normalizer.