We already know the theorem
Theorem
Let $p: (Y,y) \rightarrow (X,x)$ be a covering, with $Y$ connected and $X$ locally path connected, and let $p(y) = x$. If $p_*(\pi_1(Y,y))$ is a normal subgroup of $\pi_1(X,x)$, then $\pi_1(X,x)/p_*(\pi_1(Y,y))$ is isomorphic to Aut(Y/X).
Question
As the situation above, T.F.A.E
(a) the covering is normal (i.e., $p_*(\pi_1(Y,y))$ is a normal subgroup of $\pi_1(X,x)$);
(b) the action of $Aut(X/Y)$ on $p^{-1}(x)$ is transitive;
(c) for every loop $\sigma$ at $x$, if one lifting of $\sigma$ is closed (i.e., still a loop at some $y \in p^{-1}(x)$), then all liftings are closed.
My approach
I already proved (b) $\implies$ (c) and (c) $\implies$ (a). So I need to prove (a) $\implies$ (b), and I know I need to use the theorem above.
So, is that true that for any $y$ and $y' \in p^{-1}(x)$, the there exists $[\sigma] \in \pi_1(X,x)$ such that $y' = [\sigma]y$? (If this is true, then we are done!)
Thank you very much!