Let $G$ be a group and $I$ be a normal subgroup of $G$ such that $G/I$ is cyclic, and let $\bar{g}$ be a generator. Let $g, g' \in G$ be two lifts of $\bar{g}$, i.e. $g + I = g' + I = \bar{g}$.
Question: Are $g$ and $g'$ conjugated, i.e. does there exist an element $h \in G$ such that $g' = hgh^{-1}$?
Approach: If we would have a surjective group homomorphism $\varphi: G \to H$ where $H$ is cyclic, then $G/I \simeq H$ where $I = \ker{\varphi}$. If $\bar{g}$ is a generator of $G/I$ and $g$, $g' \in G$ be two lifts of $\bar{g}$, then $\varphi(m g m^{-1}) = \varphi(g) = \varphi(g')$ for all $m \in I$. So it would be neat if it is possible to find an element $m \in I$ such that $mgm^{-1} = g'$. But I don't know if that is going to be useful or possible.
Could you please help me with this problem? Thank you!