Fixed points of finite group I was wondering if we have a finite group $G$ with subgroups $H$ and $K$.  With $X:=G/H$.
Is there a simple way of finding $X^K$ and when will $X^K=\emptyset$? 
Also does this have any baring on when $G/H≅ G/K$?
(This is the first question I've ever asked here so apologies for errors etc.)
 A: I'm assuming that $G/H$ represents the left cosets of $H$ in $G$, that is the sets of the form $gH$ with $g\in G$, with the action of $G$ on $X$ being given by $x\cdot(gH) = xgH$ (left multiplication), and that $X^K$ is the set of fixed points of $K$; that is, all $gH\in X$ such that for all $k\in K$, $k(gH)=gH$.
If $kgH = gH$, then $g^{-1}kg\in H$. Thus, $gH$ is fixed by all $k\in K$ if and only if $g^{-1}Kg\subseteq H$. So
$$X^K = \{ gH\in G/H \mid g^{-1}Kg\subseteq H\}.$$
Note that this is well defined: if $gH = g'H$, then $g'^{-1}g\in H$, so if $g^{-1}Kg\subseteq H$, then 
$$g'^{-1}Kg' = (g'^{-1}g)(g^{-1}Kg)(g'^{-1}g)^{-1}\subseteq g'^{-1}gH(g'^{-1}g)^{-1}=H.$$
That gives an easy way to find $X^K$ (by looking at the conjugates of $K$) and to figure out if $X^K=\emptyset$ (if no conjugate of $K$ is contained in $H$; e.g., if the order of $K$ is larger than the order of $H$; if $H$ is normal and $K$ is not contained in $H$; many other criteria).
Now, when are the $G$-sets $G/H$ and $G/K$ isomorphic? First, you must have $[G:H]=[G:K]$. Also, there must be a bijection $\psi\colon G/H\to G/K$ such that for all $x,g\in G$, $\psi(gxH) = g\psi(xH)$. In particular, for all $h\in H$ we must have $\psi(H) = \psi(hH) = h\psi(H)$. If $\psi(H) = xK$, then that means that $HxK = xK$, or that $x^{-1}Hx\subseteq K$. In particular, we need a conjugate of $H$ to be contained in $K$. Since $G$ is finite and $H$ and $K$ must have the same order, that means that $H$ is a conjugate of $K$. If $x^{-1}Hx=K$, then for any coset $gH$ we must have $\psi(gH) = g\psi(H) =gxK$. 
Conversely, if $H$ is a conjugate of $K$, $x^{-1}Hx = K$, then let $\psi\colon G/H\to G/K$ be the map that sends the coset $gH$ to the coset $gxK$. First, I claim this is well defined: if $yH=zH$, we need to show that $yxK = zxK$. Since $z^{-1}y\in H$, then $x^{-1}z^{-1}yx\in x^{-1}Hx = K$, so $yxK=zxK$, as claimed. If $gH\in G/H$ and $a\in G$, then $\psi(a\cdot gH) = \psi(agH) = agxK = a\cdot gxK = a\psi(gH)$, so $\psi$ is a $G$-set homomorphism. The map $G/K\to G/H$ given by $gK\mapsto gx^{-1}K$ is also a $G$-set homomorphism (as $xKx^{-1}=H$), and is the inverse of the previous map, so $\psi$ is an isomorphism. Thus, the $G$-set $G/H$ is isomorphic to the $G$-set $G/K$ if and only if $H$ and $K$ are conjugate. 
Added: It is interesting to note that the definition of $\psi$ above does depend on the choice of $x$; if $x$ and $y$ are two distinct elements such that $xHx^{-1}=yHy^{-1}=K$, then the maps $\psi(gH) = gxK$ and $\phi(gH) = gyK$ are the same map if and only if $y^{-1}x\in K$, that is, if and only if $y$ and $x$ are in the same coset of $K$. So you get one map for each coset of $K$ that intersects the set of elements that conjugate $H$ to $K$. 
Note, however, that this does not give you any light into the situation in which $H$ and $K$ are normal, and you are interested in knowing whether the quotient groups $G/H$ and $G/K$ are isomorphic as groups: obviously, if $H$ and $K$ are normal, then they are conjugate if and only if they are equal, but it is trivial to find examples of two distinct normal sugroups of a finite group $G$ that give you isomorphic quotients. For example, take any two distinct nontrivial proper subgrups of the Klein $4$-group.
