I proved the following proposition as an exercise:
Suppose $H \leq K \leq G$ are groups and that $G$ acts on $\frac{G}{H}$ and $\frac{G}{K}$. If $H$ is subconjugate to $K$ (i.e., if $\exists g \in G$ such that $g^{-1}Hg \leq K$), then there exists a $G$-equivariant function $f:\frac{G}{H} \rightarrow \frac{G}{K}$.
I defined $f$ by mapping $\gamma H$ to $(\gamma g) K$; well-definedness and equivariance were pretty routine to check.
However, I am told that the converse is also true; i.e. that if I'm given an equivariant function $f$, then $H$ is subconjugate to $K$. I don't see how to prove it in this direction (and I'm feeling rather shaky with group actions in general) so any help would be appreciated!
Suppose there is a function $G$-equivariant $f:G/H \to G/K$, then $$ f(gH) = g\cdot f(H) \quad\forall g\in G $$ Now let $g_0\in G$ such that $f(H) = g_0K$, then for any $h\in H$, one has $$ g_0K = f(H) = f(hH) = h\cdot f(H) = hg_0K $$ and so $g_0^{-1}hg_0 \in K$ for all $h \in H$, and hence $g_0^{-1}Hg_0 \subset K$