Let $G$ be a group , $X=\{f:G\to \{ 1,\cdots ,n\}\}$
Prove the action $\forall g\in G , f\in X , (g\cdot f)(x):=f(g^{-1}x)$ is group action.
First, $$\begin{align} (e\cdot f)(x)&=f(e^{-1}x)\\ &=f(ex)\\ &=f(x). \end{align}$$
Second, $$\begin{align} (gh\cdot f)(x)&=f((gh)^{-1}x)\\ &=f(h^{-1}g^{-1}x)\\ &=f(h^{-1}(g^{-1}x))\\ &=h\cdot f(g^{-1}x)\\ &=h(g\cdot f)(x). \end{align}$$
Is my solution sufficient (and correct) ? Am I suppose to prove that the action is well-defined? how am I suppose to prove it ?
Any help is welcome.
You've made a mistake in part (2). You want to prove that $(gh) \cdot f = g \cdot (h \cdot f)$.
Think of the function which takes $y \in G$ and gives $f( h^{-1} y)$. This is the function $h \cdot f$.
So $(g \cdot (h \cdot f))(x) = (h \cdot f)(g^{-1} x) = f (h^{-1} (g^{-1} x))$, which is the same expression you had for $((gh) \cdot f)(x)$.
Other than that, your proof looks good. There is no ambiguity (or arbitrary choice) in the definition of the action, so there is no need to check that anything is well-defined.