What is this group explicitly? Let $G$ be finite group act on a set $X$ transitively. I already proved the set $\{ f : X \to X | f(g*x) = g*f(x) \, \forall x \in X, g \in G \}$ is a group. 
My question is what is this group explicitly in terms of $G$? I know the center of $G$ will be in that set but it seems that set contains more and I can't get further.
Any help is appreciated.
 A: You are looking at the set of $G$-set endomorphisms of $X$. An example of where this could be bigger than the group is as follows. Let $G$ be the group of rotations of $\mathbb{C}$ generated by $e^{\frac{2\pi}{n}}$ with $n>1$ an integer. Let $X=\mathbb{C}$. Then maps of the form $f(z)=\alpha\cdot z$ are $G$-set endomorphisms for every complex $\alpha\neq0$. 
A non-finite example: Let $G$ be the set of invertible matrices with entries in $\mathbb{Q}$ acting on a $n$-dimensional vector space over $\mathbb{Q}(\sqrt{2})$. Then the set you are trying to describe includes the map swapping $\pm\sqrt{2}$ and fixing rationals, which is not an element of your group.
Sometimes, the set can be put in one to one correspondence with $G$.
I hope this helps.
A: Let $G$ be a group. The $G$-sets together with $G$-equivariant morphisms consistute a category. For every object of a category, we may consider its automorphism group. Specifically, if $X$ is a $G$-set, its automorphism group consists if bijective $G$-equivariant maps $X \to X$. (What I want to say here is that there is no need to give a special proof that this is a group, since this is true by general nonsense.)
Every transitive $G$-set is isomorphic to $G/H$ for some subgroup $H$ of $G$ (namely, any stabilizer subgroup of a point). So to understand automorphisms of transitive $G$-sets, we only have to consider automorphisms of $G/H$. Since $g = g \cdot 1$ in $G$ and hence $gH = g \cdot 1H$ in $G/H$, we see that any homomorphism of $G$-sets $G/H \to G/H$ is completely determined by the image of $1H$. If $xH \in G/H$, there is at most one homomorphism $G/H \to G/H$ mapping $1H \to xH$, namely $gH \mapsto gxH$. But this is well-defined iff $gH=g'H$ implies $gxH=g'xH$ for all $g,g'$ iff (exercise..) $x^{-1} Hx=H$, i.e. $x \in N(H)$ (normalisator). Notice that the map is automatically an isomorphism with inverse induced by $x^{-1} H$, so that here the automorphism group consists of all endomorphisms. We conclude:
$$\mathrm{Aut}(G/H) \cong N(H)/H$$
This group, of course, depends on the subgroup $H$. The two extreme cases are: (1) $H$ is normal, then $\mathrm{Aut}(G/H) \cong G/H$.  (2) $N(H)=H$ (for example when $H$ is not normal but maximal), then $\mathrm{Aut}(G/H) \cong \{1\}$.
