$H \triangleleft G$, $h \in H$. $C_G(h)g \cap H \ne \emptyset, \forall g \in G$ implies $\lbrace C_G(h)g \cap H, g \in G \rbrace$ partition of $H$?

Question

Let $G$ be a group and $H \triangleleft G$. I know that $G$ fixes the conjugacy classes of $H$ under conjugation if and only if $\forall h \in H, C_G(h)g \cap H \ne \emptyset, \forall g \in G$.

Now, $\forall h \in H$, call $R_g(h):=C_G(h)g \cap H$. I'm arguing that, perhaps under stronger hypothesis than above, $R:=\lbrace R_g(h), g \in G \rbrace$ is a partition of $H$, different than the one induced by conjugacy (namely $O=\lbrace O_h, h \in H \rbrace$, $O_h$ being the orbit by $h$).

Is that it?