Burnsides' Lemma for powers of fixed ponts I am familiar with Burnside's lemma:
$$\frac{1}{|G|}\sum_{g \in G} |\text{Fix}(g)| = |X / G|$$
where $X/G$ denotes the number of orbits of $X$ and ${\rm Fix}(g)$ denotes the set of fixed points of $G$. However, I came across this post where it is answered that
$$\frac{1}{|G|}\sum_{g \in G} |\text{Fix}(g)|^n$$
is the number of orbits of $G$ acting on $X^n$. This seems obvious to me, but I frustratingly can't seem to prove it. I'm thinking of trying an inductive method since Burnside's lemma proves the $k=1$ case, but I get stuck when considering the $k+1$ case with $\frac{1}{|G|}\sum_{g \in G}|\text{Fix}(g)|^{k+1}$.
 A: A point in $X^n$ is a fixed iff each coordinate is fixed. Writing $X^g$ for the fixed points of $g\in G$ under the action $G\curvearrowright X$, we can phrase this as $(X^n)^g=(X^g)^n$ (as sets).
Burnside's says $\displaystyle|X^n/G|=\frac{1}{|G|}\sum_{g\in G}|(X^n)^g|$, so just rewrite the summand.
A: As a start, I'd take it easier with $n=2$. The $G$-action on $X$, say "blank" (no infix symbol), induces a $G$-action on $X\times X$, say "$\star$", via:
$$g\star(x_1,x_2):=(gx_1,gx_2) \tag 1$$
By definitions of "$\operatorname{Fix}$" and equality between $2$-tuples, and by $(1)$, we get:
\begin{alignat}{1}
\operatorname{Fix}_\star(g) &= \{(x_1,x_2)\in X\times X\mid g\star(x_1,x_2)=(x_1,x_2)\} \\
&= \{(x_1,x_2)\in X\times X\mid gx_1=x_1\wedge gx_2=x_2\} \\
&= \{(x_1,x_2)\in X\times X\mid x_1\in\operatorname{Fix}(g)\wedge x_2\in\operatorname{Fix}(g)\} \\
&= \operatorname{Fix}(g)\times \operatorname{Fix}(g)
\end{alignat}
whence:
$$\left|\operatorname{Fix}_\star(g)\right|=\left|\operatorname{Fix}(g)\right|^2$$
But "$2$" doesn't play anything special here, so indeed:
$$|X^n/G|=\frac{1}{|G|}\sum_{g\in G}\left|\operatorname{Fix}_\star(g)\right|=\frac{1}{|G|}\sum_{g\in G}\left|\operatorname{Fix}(g)\right|^n$$
