Counterexample for the normalizer being a normal subgroup

Let $G$ be a group, and $H$ a subgroup.

$H$'s normalizer is defined: $N(H):=\{g\in G| gHg^{-1}=H \}$.

Prove $N(H)$ is a normal subgroup of G, or give counterexample.

Intuitively it seems to me that this claim is wrong, however, I'm having trouble with finding a counterexample.

Thans in advance for any assistance!

The normaliser of any proper nontrivial subgroup of a simple group would work too. Can you see why?

An easy, concrete counterexample is any of the three subgroups of order $2$ in $S_3$.

For any finite group $G$ and any $p$-Sylow subgroup $P \leq G$ the following holds:

Every subgroup $U$ with $N_G(P) \leq U \leq G$ satisfies $N_G(U) = U$.

Especially every normalizer $N_G(P)$ provides a counterexample unless $P$ is normal in $G$. Can you think of examples of Sylow subgroups which are not normal?