Let $G$ be a finite group and let $H \leq G$ be a subgroup. Furthermore, let $G$ act on $G/H$ by left multiplication, i.e. $g * sH := g s H$.
I would like to show that the number of fixed points of an arbitrary element $g$ is $$ \dfrac{|Cl(g) \cap H| \cdot |C_G (g)|}{|H|}, $$ where $Cl(g) = \{s g s^{-1} : \tilde{g} \in G \}$ is the conjugacy class of $g$, and $C_G (g) = \{s \in G : sg = gs \}$.
So far, it is clear to me that if $sgs^{-1} \in H$, then $sH$ is a fixed point of $g$, so I can see where the $|Cl(g) \cap H|$ term might come from. Otherwise I'm not sure how to show this.
Thanks a lot.
Consider the map $\varphi:G\longrightarrow G$ given by $x\longmapsto x^{-1}gx$: now, $xH$ is fixed by $g$ iff $x^{-1}gx$ belongs to $H$. That is to say, iff $x\in \varphi^{-1}(\text{Cl}(g)\cap H)$.
Now, it is easy to see that $|\varphi^{-1}(\text{Cl}(g)\cap H)|$ equals $|\text{Cl}(g)\cap H|\cdot|C_G(g)|$, by the fact that $C_G(g)$ is the stabilizer of $g$ in the action of $G$ on itself by conjugation.
Therefore, the number of elements $x\in G$ such that $gxH=xH$ is $|\text{Cl}(g)\cap H|\cdot|C_G(g)|$: since you wish to count the cosets of $H$ for which the same equation holds, you need to divide it by $|H|$.