Given topological spaces $X$ and $Y$ and a covering map $p: X \rightarrow Y$, we know that the group $\pi_1(Y,y_0)$, where $y_0\in Y$, acts on the fiber $F=p^{-1}(y_0)$.
Also, we know that the set of all action $G\times M \rightarrow M$ of a group $G$ on a set $M$ is in a bijective correspondence with the set of all group homomorphisms $G\rightarrow \text{Sym}(M)$. Here $\text{Sym}(M)$ is the symmetric group on the elements of $M$.
In particulat, the action of the fundamental group of $X$, as any action of a group on a set, by the above, induces the group homomorphism $\pi_1(Y,y_0)\rightarrow \text{Sym}(F)$. Similarly, by the above, any group homomorphism $\pi_1(Y,y_0)\rightarrow \text{Sym}(F)$ must induce the action of $\pi_1(Y,y_0)$ on the fiber $F$. But I was told that this is not true. The question is why? In such a case, what additional requirements one must impose on the homomorphism $\pi_1(Y,y_0)\rightarrow \text{Sym}(F)$ in order that it induce the action on the fiber? Is it true that the homomorphism must be transitive? Does it have something to do with the Riemann Existence theorem? (If you refer to this theorem, please use its statement presented here.)