Let $G$ be a finite group.
Let $C$ be the category with objects subgroups of $G$ and morphisms between two subgroups $H,H'$ be $ \{ g \in G \mid g H g^{-1} \subset H ' \}$.
Let $D$ be the category with objects are pairs: $X$ a set, and a transitive $G$ action on $X$. The morphisms between two pairs $X,X'$ and their $G$ actions is a function $f: X \to X'$ such that $ f(g*x ) = g* f(x) $ for any $ x \in X, g \in G$.
I want to show that there is a full essentially surjective functor from $C$ to $D$.
I think the functor should map any subgroup $H$ into the cosets $ G / H$ with the left multiplication as the group action. (Since if I do this the functor will easily being essentially surjective).
But I have trouble on mapping the morphisms. For any $g $ such that $g H g ^{ -1} \subset H '$, I think I should map it to the function from $G/ H \to G/ H'$ by sending $ g' H $ to $ gg' g^{ -1 } H ' $. Since if I do this the function will easily check to be well-defined.
But if I choose to map the morphisms as above, I have trouble showing that the morphisms I map to is actually a morphism in $D$, i.e. I don't know how to show for any $g', g'' \in G$, two cosets $g g'' g' g^{ -1 } H ' = g'' g g' g^{-1} H ' $ are equal.
I tried a long time and now I have totally no confidence if my construction works. Any help is appreciated.
This is sort of a rough idea:
What you have just said is this:
$C(H,H')=\{{\lambda} \in G : {\lambda}H{\lambda}^{-1} \subset H \}$
You have defined a functor $F$ which has object function which will send $H \mapsto G/H$. You want this functor to be "full" or surjective on hom sets.
In this case our corresponding hom set to $C(H,H')$ is $D(G/H, G/H')$
If you choose $f \in D(G/H, D/H')$ then we have $f(eH)={\lambda}H'$.
Which implies ${\lambda}H'=f(eH)=f(hH)=hf(eH)=h{\lambda}H'$. Which implies that ${\lambda}^{-1}h{\lambda} \in H'$ for all $h \in H$.
I hope this helps.