In Peter Scott's paper "the geometries of 3-manifolds" he has a Lemma 3.5 which describes the naturality property of the obstruction term $b$ under a finite cover. This is given by $b(\bar{\eta})=b(\eta)\frac{l}{m}$, where $l$ is the order of the cover of the orbit spaces and $m$ is the order that fibers of $\bar{\eta}$ cover fibers of $\eta$.
Suppose a finite group $G$ acts fiber-preservingly on some $M$ which is an orientable Seifert manifold without critical fibers and obstruction $b$. Suppose further that the induced action on the (orientable) orbit space is free. So $b=b(M/G)\frac{l}{m}$. If $m \ne 1$, is there some other finite group $G'$ that acts fiber-preservingly on $M$ which has the same induced action on the orbit space as $G$, but $m'=1$?
My goal is to show that if the induced action $\hat{G}$ on the orbit space is free, then the order of $\hat{G}$ divides the obstruction term $b$.
EDIT: The result follows if $l$ and $m$ are coprime.