Let $H$ be a subgroup of a finite group $G$, and let $N = N_G(H)$ be the normalizer of $H$ in $G$.
For $x \in G$ is there a lower bound for $[ H : H \cap xHx^{-1} ]$? If $x \in N$ this index is 1, of course. If $x \notin N$ do we have $[N : H] \leq [H : H \cap xHx^{-1} ]$? What if $N/H$ is cyclic?
You can make $[N:H]$ as big as you want: start with an arbitrary group $E$ and non-normal subgroup $H$ (e.g. $E=C_p\rtimes C_2$) dihedral of order twice a prime $p$, and $H=C_2$; for every $x\not\in H$ one has $[H:H\cap xHx^{-1}]=2$), and take $G$ to be the direct product of $E$ and any other group $U$, cyclic if you want. Then in $G$, the normaliser of $H$ will contain $U$, so $[N:H] \geq |U|$. If in the above example you take $U$ to be cyclic of order gazillion, then $N=UH$, and $N/H$ will be cyclic of order gazillion.
Edit to address the modification in the comment: merely demanding that the extension of $N/H$ by $H$ is non-split is not enough to eliminate "silly" example as above: take a group $X$ whose automorphism group contains $S_3$, the symmetric group of order $6$, and such that $C_3\leq S_3$ acts fixed point freely on $X$ (e.g. $X=C_2\times C_2$ will do), and take the semidirect product $E=X\rtimes S_3$. Let $H$ be the subgroup of order $3$ in $S_3$. Then for every $x\in E$ that does not normalise $H$ one has $[H:xHx^{-1}]=3$. Now take the direct product $G$ of $E$ and any cyclic group $U$ of odd order. Then the normaliser of $H$ in $G$ is generated by $S_3$ and $U$. Since $U$ is cyclic of odd order, $N/H\cong C_2\times U$ is cyclic, but now the $C_2$ acts non-trivially on $H$, so the extension is not split.
