I'm seeking a normal subgroup $H$ of a group $G$ such that for some element $a \in G$, we have that $aHa^{-1} \subset H$ yet not $aHa^{-1} = H$.
Yet it seems to me this is impossible since $|aHa^{-1}| = |H|$ since if $f: H \rightarrow aHa^{-1}$ s.t. $f(h_i) = ah_ia^{-1}$, then $f$ is surjective (any $ah_ia^{-1} \in aHa^{-1}$ gets mapped to by $f(h_i)$) and $f$ is injective (if $h_i \ne h_j$ then $ah_ia^{-1} \ne ah_ja^{-1}$ since to assume otherwise would imply $a^{-1}(ah_ia^{-1})a = a^{-1}(ah_ja^{-1})a$ so that $h_i = h_j$ absurdly).
How -- given that such a bijection $f$ exists between $H$ and $aHa^{-1}$ -- could it be that there exists a normal subgroup $H$ of $G$ s.t. $aHa^{-1}$ is a proper subset of $H$?
EDIT: In fact, for any normal $H$ and for any $a \in G$, we have that $aHa^{-1} = Haa^{-1} = H$ so that it seems fundamentally impossible for this reason alone that $aHa^{-1} \subset H$. Here it doesn't seem to matter if $H$ is finite or infinite, doesn't it?