Let $G$ be a group, and let $H$ be a subgroup of finite index in $G$, and let $N \colon = \cap_{x \in G} \ xHx^{-1}$. Then $N$ is clearly a subgroup of $G$ which is contained in $H$ and such that $aNa^{-1} = N$ for all $a$ in $G$.
Is $N$ of finite index in $G$? And If so, what can we say about an upper bound for the index of $N$? If not, then can we find a subgroup $N^\prime$ of finite index in $G$, contained in $H$, and such that $aN^\prime a^{-1} = N^\prime$ for all $a$ in $G$? And what can we say about the index of $N^\prime$ in $G$?
To your questions: yes, and an upper bound is $\;[G:H]!\;$ (the factorial of the index).
Short explanation: if we denote by $\;X\;$ the set of left cosets of $\;H\;$ in $\;G\;$, we make $\;G\;$ act on $\;X\;$ by left shift: $\;x\cdot gH:=(gx)H\;,\;\;g,x\in G\;$.
The above action determines a homomorphism $\;\phi:G\to\text{Sym}_X\cong S_n\;,\;\;n=[G:H]\;$, and it happens to be that $\;\ker\phi=\cap_{g\in G}H^x\;$ is the maximal subgroup normal in $\;G\;$ that is contained in $\;H\;$ .
We also have, of course, that $\;\ker\phi\lhd G\;$ and by the above $\;[G:\ker\phi]\mid |S_n|=n!\;$.
The subgroup $\;\ker\phi\;$ described above is called the core of $\;H\;$.