Let $1 \mapsto A \mapsto B \mapsto C \mapsto 1$ be a short exact sequence of groups and assume that $B$ is finitely generated.
My question is whether the action of $C$ on $A$ (by conjugation) has finitely many orbits.
If $C$ is finitely presented, I think that it is true. Since $C$ is finitely presented, $A$ is finitely generated as a normal subgroup of $B$, in other words, the action of $C$ on $A$ has finitely many orbits.
Nevertheless, if we do not ask $C$ to be finitely presented (we just know that since $B$ is finitely generated, so is $C$), I do not know whether the following argument is correct:
The group $B$ is finitely generated and it is generated by $\langle A, \tilde{C} \rangle$, where $\tilde{C}$ is a lift of $C$ in $B$. Then, there are elements $a_1,\dots,a_k$ in $A$ and $\tilde{c_{1}},\dots,\tilde{c_{n}}$ in $\tilde{C}$ such that $B=\langle a_1,\dots,a_k,\tilde{c_{1}},\dots,\tilde{c_{n}}\rangle$. Hence, $A$ is generated by $a_1,\dots,a_k$ and their conjugates by words in $\{\tilde{c_1},\dots,\tilde{c_{k}}\}$. In particular, the action of $C$ on $A$ has finitely many orbits because $$A=\bigcup_{i\in \{1,\dots,k\}} C \cdot a_i.$$