Counting the number of orbits with trivial symmetries Let $G$ be a finite group acting on a finite set $X$. The Lemma that is not Burnside's tells us that $$
|X/G| = \frac{1}{|G|}\sum_{g\in G} |X^g|
$$ where $|X/G|$ is the number of orbits and $|X^g|$ denotes the number of elements fixed by $g$, i.e., $X^g=\{x\in X\mid g\cdot x=x\}$.
Consider the partition $X=Y\cup Z$ where $Y=\{x\in X\mid |G\cdot x|=|G|\}=\{x\in X\mid stab_G(x)=1\}$ is the set of elements with trivial stabilizers (which is $G$-stable) and $Z$ is the complement of $Y$. Can I count the number of orbits in $Y$?
Clearly, $$|X/G|=\frac{1}{|G|}\bigg(\sum_{1\neq g\in G}|Z^g|+|Z|+|Y|\bigg) = |Z/G|+|Y/G|. $$
I do not know the size of $Z$ or $Y$, all I know are the numbers $|X|, |X^g| $ for $g\in G$ (of course I also know the group). Can I calculate $|Y/G|$ from the above equation?
 A: 
We can try to express |Y| and |Z| in terms of |X|.
Firstly note that
$\frac{1}{|G|}\sum_{g \in G} |X^g| = \frac{1}{|G|}( \sum_{g \in (G/1)} |Z^g| + |Z| + |Y|) = |Z/G| + |Y/G|$
Now, we can multiply by |G| giving us:
$\sum_{g \in G}|X^g| = \sum_{g \in (G/1)}(|Z^g| + |Z| + |Y|) = |G||Z/G| + |G||Y/G|$
From here, we can see that since in the case of g = 1, $|X^g|$ = |X|,
$\sum_{g \in (G/1)}|X^g| + |X| = \sum_{g \in (G/1)}(|Z^g| + |Z| + |Y|) = |G||Z/G| + |G||Y/G|$
Furthermore, since  $X = Y \bigcup Z$ that |X| = |Y|+|Z|, so we can say
$\sum_{g \in (G/1)}|X^g| +|X| = \sum_{g \in (G/1)}(|Z^g| + |X|) = |G||Z/G| + |G||Y/G|$
We can now split our second sum and get $\sum_{g \in (G/1)} |X| = (|G|-1)|X|$. From there, we can also subtract |X| and say
$\sum_{g \in (G/1)}|X^g| = \sum_{g \in (G/1)}|Z^g| + (|G|-2)(|X|)= |G||Z/G| + |G||Y/G| - |X|$
Now with this expression, we can find |Z/G| by Burnside's lemma,
$|Z/G| = \sum_{g \in (G/1)}|X^g| - (|G|-2)|X| $
and now finally, we can say
$\sum_{g \in G} |X^g| - |G|(\sum_{g\in (G/1)}|X^g| - (|G|-1)(|X|)) = |G||Y/G| $
$|Y/G| = \frac{1}{|G|} \sum_{g \in G} |X^g| - \sum_{g\in (G/1)}|X^g| - (|G|-1)(|X|)$
I hope this helped, and please point out any errors I've made :)
A: We unfortunately cannot recover such detailed information from the cardinalities of the fixed points, for groups that arent cyclic. For a concrete counterexample, consider $G=S_3$, and consider the $G$ sets $$S_3/C_3\coprod S_3/C_2\coprod S_3/C_2$$ and $$S_3/e\coprod S_3/S_3 \coprod S_3/S_3.$$
They have the same number of fixed points for all $g\in G$, but only one has nontrivial free locus.
To understand whats going on, we can interpret this representation theoretically. For any finite group, we can consider the Burnside ring $B[G]$ of $G$ sets, the ring of isomorphism classes of (virtual) $G$ sets, under disjoint union and product. This has a natural homomorphism (taking the free vector space) to the representation ring $R[G]$, (say over $\mathbb{C}$), which is the same construction, except applied to representations of $G$ under direct sum and tensor product. Ordinary character theory says we have an isomorphism $$R[G]\otimes_\mathbb{Z} \mathbb{C}\cong \text{Fun}_G(G)$$ given concretely by sending a representation to its character, a $\mathbb{C}$ valued class function on $G$.
We may observe that the the class function obtained from a $G$ set given by looking at fixed points is precisely the character of the associated representation, so by looking at fixed points of elements only, we will be able to detect the representation associated to the $G$ set, but not necessarily the $G$ set itself. The reason we can count orbits is that this information is visible on the representation level, its given by the dimension of the space $\text{Hom}_G(\mathbf{1},\mathbb{C}[X])$ where $X$ is our $G$ set.
If you instead look at the number of fixed points for all conjugacy classes of subgroups, you can actually recover the $G$ set, and we have a similar "character isomorphism" $$B[G]\otimes_\mathbb{Z} \mathbb{Q} \cong \{f:\text{Conjugacy classes of subgroups}\rightarrow \mathbb{Q}\}.$$
