Is there formulas for $\sum_{g\in G}|\operatorname{Fix}(g)|^n$? I saw a problem in Show $r(x)|G|=\sum_{g\in G}|Fix(g)|^2$ where $(G,X)$ is a transitive group and $r(x)$=#{different orbits of X under $Stab(x)$}
Let $(G,X)$ be a transitive group action, that is, for any $x,y\in X$,there exists an element $g\in G$ s.t. $g*x=y$. Fix $x\in X$. Let $r(x)$ be the number of different orbits of $X$ under $\operatorname{Stab}(x)$. Show that  $r(x)|G|=\sum_{g\in G}|\operatorname{Fix}(g)|^2$.
Can this be generalized such that if the assumptions stays on the same and $n\in \mathbb{N}$ then $\sum_{g\in G}|\operatorname{Fix}(g)|^n$ can be represented in some closed form with respect to the variables $r(x)$ and $|G|$? If there is not known general solution, what do we know about some fixed powers of $n$?

$|\text{Fix}(g)|^n$ is the number of fixed points of $G$ acting diagonally on $X^n$. By Burnside's lemma, it follows that
$$\frac{1}{|G|} \sum_{g \in G} |\text{Fix}(g)|^n$$
is the number of orbits of $G$ acting on $X^n$.
If $G$ acts transitively on $X$ then we can count the number of orbits of $X^2$ as follows. First, $X^2$ splits into the diagonal elements $(x, x)$, which consist of one orbit, and then the off-diagonal elements $(x, y), y \neq x$. Since $G$ acts transitively we may pick representatives of each orbit of the form $(x_0, y)$ for any fixed $x_0 \in X$. By orbit-stabilizer the number of such orbits is then the number of orbits of $y$ under the action of $\text{Stab}(x_0)$.
You can do a similar but more complicated analysis for $X^n$ by breaking its orbits up according to which elements are equal to which other elements (equivalently, breaking it up according to the orbits of the action of $\text{Sym}(X)$), which shows in particular that the above sum is at least the number of partitions of $\{ 1, 2, \dots n \}$ into at most $|X|$ nonempty subsets (this is a sum of Stirling numbers of the second kind $S(n, k), k \le |X|$), with equality iff the action of $G$ is $k$-transitive for $k \le n$. 
