Let $G$ be a group containing a finite central subgroup $Z$. Considering the conjugation action of $G$ on $G/Z$, define the stabilizer subgroups $$\operatorname{Stab}_G(gZ) = \{h \in G\colon hgh^{-1} = gz\text{ for some }z \in Z\}\,.$$ These contain the centralizers $C_G(g)$, but a priori they may differ. My question is the following:
Question If the index $|G\colon \operatorname{Stab}_G(gZ)|$ is finite, can the index $|\operatorname{Stab}_G(gZ)\colon C_G(g)|$ be infinite?
In other words, if $g \in G$ is such that the conjugacy class of $gZ$ is finite in $G/Z$, can the conjugacy class of $g$ in $G$ be infinite?
Denote by $C$ the conjugacy class of $g$ and by $D$ the class of $gZ$. Then, consider the function $$f \colon C \to D$$ defined by $f(h) = hZ$. We have that $f^{-1}(hZ) = \{k \in C \colon k = hz,\text{ for some }z \in Z\}$, therefore, the cardinality of any fiber is less than or equal to $|Z|$. Since $D$ and $Z$ are finite, we conclude that \begin{align*} |C| & = \bigg|\bigcup_{hZ \in D} f^{-1}(hZ)\bigg| \\ & \leq |D|\cdot|Z| \end{align*} and, therefore, the conjugacy class of $g$ is finite.