Let $G$ be a group and $H \triangleleft G$, $H \ne \lbrace e \rbrace$. For given $h \in H$ and $g \in G$, let's set $R_g(h):=C_G(h)g \cap O_h$, where $C_G(h)$ is the centralizer of $h$ in $G$, and $O_h$ is the orbit through $h$ by conjugacy. We get:
$$\begin{align} R_g(h) \ne \emptyset &\Leftrightarrow \exists y \in C_G(h), \exists x \in G \mid yg=x^{-1}hx \\ & \Leftrightarrow \exists z,x \in G \mid (zg=x^{-1}hx) \wedge (zh=hz) \\ &\Leftrightarrow \exists z,x \in G \mid (hzg=hx^{-1}hx) \wedge (zh=hz)\\ &\Leftrightarrow\exists z,x \in G \mid (zhg=hx^{-1}hx) \wedge (zh=hz)\\ &\Leftrightarrow (z=hx^{-1}hxg^{-1}h^{-1}, x \in G) \wedge (zh=hz). \end{align}$$
Now, $x \in C_G(h) \Rightarrow z= h^2g^{-1}h^{-1}$; but $h^2g^{-1}h^{-1} \in C_G(h) \Leftrightarrow g \in C_G(h)$, so that:
- $g \in C_G(h) \Rightarrow R_g(h)=\lbrace h\rbrace \cup \lbrace zg=hx^{-1}hxg^{-1}h^{-1}g \mid (x \in G \setminus C_G(h)) \wedge (zh=hz) \rbrace$
- $g \in G \setminus C_G(h) \Rightarrow R_g(h)=\lbrace zg=hx^{-1}hxg^{-1}h^{-1}g \mid (x \in G \setminus C_G(h)) \wedge (zh=hz) \rbrace$
Special case: $G$ abelian. Then, $C_G(h)=G$, $O_h=\lbrace h \rbrace$, $R_g(h)=\lbrace h \rbrace$.
Can $\lbrace zg=hx^{-1}hxg^{-1}h^{-1}g \mid (x \in G \setminus C_G(h)) \wedge (zh=hz) \rbrace$ be simplified somehow?