Locus equation in a non-simple group_part#2 (This is more a focused post than this similar one of mine, and the question more direct.)
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. Then: 
\begin{align}
R_g(h) &= \lbrace x^{-1}hx \in C_G(h)g \mid x \in G \rbrace \\
&= \lbrace x^{-1}hx \in C_G(h)g \mid x \in C_G(h) \rbrace \cup \lbrace x^{-1}hx \in C_G(h)g \mid x \in G \setminus C_G(h) \rbrace \\
&= \lbrace h \in C_G(h)g \rbrace \cup \lbrace x^{-1}hx \in C_G(h)g \mid x \in G \setminus C_G(h) \rbrace  
\end{align}
Now, $h \in C_G(h)g \Leftrightarrow g \in C_G(h)$; so, by defining $$\tilde R_g(h):=\lbrace x^{-1}hx \in C_G(h)g \mid x \in G \setminus C_G(h) \rbrace,$$ we get:


*

*$g \in C_G(h) \Rightarrow R_g(h)=\lbrace h\rbrace \cup \tilde R_g(h)$

*$g \in G \setminus C_G(h) \Rightarrow R_g(h)=\tilde R_g(h)$

Is there any class of $G$s such that:


*

*$g \in C_G(h) \Rightarrow \tilde R_g(h) = \emptyset$

*$g \in G \setminus C_G(h) \Rightarrow |\tilde R_g(h)|=1$ ?


(Given $h$, all the [right] cosets of the centralizer of $h$ would cross the orbit by $h$ one time, and only one.)
 A: Your question can be rephrased as

For which $G$ is $|R_g(h)|=1$ for all $g,h\in G$?

or more generally

For which pairs $(H,G)$ with $H\trianglelefteq G$ is $|R_g(h)|=1$ for all $g\in G$, $h\in H$?

This can be answered completely as follows:

Lemma: If $h\notin Z(G)$ then $|R_g(h)|\ne 1$ for some $g\in G$. 

Proof: We show this by contrapositive.
Suppose $|R_g(h)|= 1$ for all $g\in G$.
From this we show that $G=C_G(h)O_h$. Indeed for $x,y\in C_G(h)$ and $z\in O_h$ there is some $x'\in C_G(h)$ with $xy=x'z$ if and only if $x^{-1}x'y=z$ which holds if and only if $z\in O_h\cap C_G(h)y=R_y(h)$. But $y\in R_y(h)$ and $|R_y(h)|=1$ so this holds if and only if $z=y$. Hence $xy=x'z$ for some $x'\in C_G(h)$ if and only if $y=z$ and therefore $x=x'$. Hence $|C_G(h)O_h|=|C_G(h)||O_h|=|G|$ by orbit-stabiliser, so $G=C_G(h)O_h$.
Now consider the natural quotient map $\varphi:G\to G/C_G(h)$. We have $\varphi(O_h)=\varphi(G)=G/C_G(h)$. But $O_h=\{h^x|x\in G\}$ and for $x\in G$ we have $\varphi(h^x)=\varphi(h)^{\varphi(x)}=1$ so $\varphi(O_h)=1$. Hence $G/C_g(h)=1$ and $h\in Z(G)$.

Corollary: Let $H\trianglelefteq G$. Then $|R_g(h)|=1$ for all $g\in G$, $h\in H$ if and only if $H\le Z(G)$. 

