(Old question and many answers already, but adding a very natural and elementary example not yet mentioned.)

Take the permutation group $\newcommand{\Z}{\mathbb{Z}}\newcommand{\Sym}{\operatorname{Sym}}\newcommand{\Stab}{\operatorname{Stab}_{\mathrm{pw}}}\Sym(\Z)$, with its subgroups the pointwise stablisers $\Stab(\Z_{<0}) < \Stab(\Z_{\leq 0}) < \Sym(\Z)$. These are evidently conjugate via the permutation “+1”.

Of course, nothing here is specific to $\Z$: it just needed a set $X$ (necessarily infinite) with an automorphism $s : X \to X$ and a subset $Y$ such that $s(Y) \subsetneq Y$; then we have $\Stab(s(Y)) < \Stab(Y) < \Sym(X)$, and they are conjugate via $s$.

More generally, $X$ can be an object of any category, $s$ an automorphism, and $Y$ a subobject of $X$ with $s Y \lneq Y$; then $\Stab(s(Y)) < \Stab(Y) < \operatorname{Aut}(X)$, conjugate via $s$ as before. Jonas Meyer’s answer is a special case of this.

(Old question and many answers already, but adding a very natural and elementary example not yet mentioned.)

Take the permutation group $\newcommand{\Z}{\mathbb{Z}}\newcommand{\Sym}{\operatorname{Sym}}\newcommand{\Stab}{\operatorname{Stab}_{\mathrm{pw}}}\Sym(\Z)$, with the subgroups of permutations acting trivially on the non-negative and the strictly positive integers respectively, $\Stab(\Z_{\geq 0}) < \Stab(\Z_{> 0}) < \Sym(\Z)$. These are evidently conjugate via the permutation “+1”.

Of course, nothing here is specific to $\Z$: it just needed a set $X$ (necessarily infinite) with an automorphism $s : X \to X$ and a subset $Y$ such that $s(Y) \subsetneq Y$; then we have $\Stab(s(Y)) < \Stab(Y) < \Sym(X)$, and they are conjugate via $s$.

More generally, $X$ can be an object of any category, $s$ an automorphism, and $Y$ a subobject of $X$ with $s Y \lneq Y$; then $\Stab(s(Y)) < \Stab(Y) < \operatorname{Aut}(X)$, conjugate via $s$ as before. Jonas Meyer’s answer is a special case of this.