How do I show such a correspondence? I know that a group action is a map $\phi: G\times X\to X$ defined by $\phi(g,x)=gx$. Any hints to connect this to the homomorphism?
Edit:
Following AOrtiz's idea, I defined $\psi:G\to S_{X}$ by $\psi(g)=\phi(g,x)$. However, I am struggling to show it is one-to-one.
My work:
Let $\psi(g_{1})=\psi(g_{2})$. Then $\phi(g_{1},x) =\phi(g_{2},x)$ or $g_{1}x=g_{2}x$. There is where I am stuck. How do I show that $g_{1}=g_{2}$?
The idea is to sort of follow your nose. Let $\psi\colon G\to S_X$ be the map defined by $g\mapsto \psi_g$, where $\psi_g(x) := \phi(g,x)$. Here $\phi\colon G\times X\to X$ is the action. You can verify that each $\psi_g$ is a bijection of $S$ by exhibiting an explicit inverse using the axioms of a group, and you can check that the map $\psi$ is a homomorphism, i.e. $\psi_{gh} = \psi_g\circ \psi_h$, by thinking about the axioms of a group action. The homomorphism $\psi$ is often called the permutation representation afforded by the action $\phi$.