Is there a $G$-equivariant bijection $h: G/X \to G/Y$? Let $X$ and $Y$ be subgroups of a group $G$ such that there are $G$-equivariant maps $f: G/X \to G/Y$ and $g: G/Y \to G/X$. Is there a $G$-equivariant bijection $h: G/X \to G/Y$?
 A: Note that there exists a $G$-equivariant map $f:G/X\to G/Y$ iff some conjugate of $X$ is contained in $Y$, and there is a $G$-equivariant bijection iff some conjugate of $X$ is equal to $Y$.  So the question is, if $gXg^{-1}\subseteq Y$ and $hXh^{-1}\supseteq Y$ for some $g,h\in G$, must $X$ be conjugate to $Y$?
The answer is no.  For instance, fix any group $X$ with subgroups $X'\subset Y\subset X$ such that $X\cong X'$ and $X\not\cong Y$ (for instance, take $X$ to be free on two generators, $Y$ any subgroup of $X$ that is free on more than two generators, and $X'$ the subgroup of $Y$ generated by two of its generators).  Let $i:X\to X$ be an injective homomorphism whose image is $X'$.  Let $L$ be the direct limit of the diagram $X\stackrel{i}\to X\stackrel{i}\to X\stackrel{i}\to X\stackrel{i}\to X\stackrel{i}\to\dots$ obtained by repeatedly appying $i$.  Then $L$ has an automorphism $\alpha:L\to L$ obtained by shifting each copy of $X$ forward (so the first copy in the diagram becomes the second, the second becomes the third, and so on).  Finally, let $G$ be the semidirect product $L\rtimes\mathbb{Z}$ where $\mathbb{Z}$ acts on $L$ via $\alpha$.  Then take the first copy of $X$ in $L$ and the copy of $Y$ inside it as subgroups of $G$.  We have $Y\subset X$, and $X$ is conjugate to the subgroup $X'\subset Y$ via $-1\in\mathbb{Z}$.  But $X$ and $Y$ are not isomorphic, so they cannot be conjugate in $G$.
