Lately I've been studying monodromy and covering maps (in particular ramified covering mapos of Riemann surfaces), and I came across something I didn't fully understand. Let $V$ be a connected real manifold, and let $\rho:\pi_1(V,q)\to S_d$ be a group homomorphism with a transitive image. Let $H=\{a\in\pi_1(V,q):\rho(a)(1)=1\}$. It is easy to prove that $H$ is a subgroup of $\pi_1(V,q)$ of index $d$. By the general theory of covering spaces, we know that there is a covering space $F:U\to V$ of degree $d$ such that $\pi_1(U,x)\cong H$.
Now, we know that $\pi_1(V,q)$ acts on $F^{-1}(q)$ by taking a class of curves $[\gamma]$ and sending a point $x$ in the fiber of $q$ to the endpoint of the lifting of $\gamma$ to $U$ with initial point $x$. This action induces a homomorphism $\pi_1(V,q)\to S_d$ (with transitive image).
Now the question: Supposedly, given a homomorphism $\rho:\pi_1(V,q)\to S_d$ with transitive image, if we take the covering $F:U\to V$ associated to the group $H$ mentioned above, why is it that the homomorphism described by the action of $\pi_1(V,q)$ on the fibers of $q$ the same (or maybe with conjugate images) as the homomorphism $\rho$?
I read this statement in Algebraic Curves and Riemann Surfaces by Rick Miranda, and can't figure out why the homomorphisms should be the same.