The number of orbits of a permutation action on subsets Let $ G $ be a subgroup of $ S_n $ acting on $ \Omega=\{ 1, \dots, n \} $. A general question is to determine the sequence $ o_k(\Omega) $ the number of orbits of the natural action of $ G $ on the $ k $-subsets. It is well known that if $ G=S_n $ then
$$
o_k(\Omega)=1
$$
for all $ k $. Similarly, for $ G=A_n $, $ n \neq 2 $ we also have
$$
o_k(\Omega)=1
$$
for all $ k $. For specific examples this can be worked out by hand. For example for the $ GL_3(2) $ subgroup of $ S_7 $ the sequence of $ o_0(\Omega), \dots, o_7(\Omega) $ is
$$
1,1,1,2,2,1,1,1
$$
Given a subgroup $ G $ of $ S_n $ is there any general method for determining the sequence $ o_k(\Omega) $?
 A: By Burnside's lemma this reduces to computing the number of fixed points of a given $g \in G$ acting on the set ${\Omega \choose k}$ of $k$-element subsets. A subset is fixed by $g$ iff it consists of a union of orbits of the action of $g$ on $\Omega$, or equivalently a union of cycles in the cycle decomposition of $g$ as a permutation. So if $c_i(g)$ denotes the number of $i$-cycles of $g$, this gives
$$o_k(\Omega) = \frac{1}{|G|} \sum_{g \in G} \prod_{\sum i k_i = k} {c_i(g) \choose k_i}.$$
This is a little messy-looking so alternatively we can argue as follows. $o_k(\Omega)$ is the dimension of the invariant subspace of the action of $G$ on $\Lambda^k(V)$ where $V = \mathbb{Q}[\Omega]$ is the free vector space on $\Omega$; this is not entirely automatic since the action of $G$ on $\Lambda^k(V)$ is not isomorphic to $\mathbb{Q}[{\Omega \choose k}]$ because of the extra signs but the extra signs turn out not to affect the argument. This means we can instead write
$$o_k(\Omega) = \frac{1}{|G|} \sum_{g \in G} \text{tr}(\Lambda^k(g))$$
which has the benefit of assembling into a generating function
$$\sum_k o_k(\Omega) t^k = \frac{1}{|G|} \sum_{g \in G} \det(1 + gt).$$
This is not an enormous simplification, though, since $\det(1 + gt) = \prod (1 + t^i)^{c_i(g)}$ so we still end up having to know the cycle decomposition of every element of the group, and if we expand this product out we just get the previous formula anyway. But at least it can save some computation to compute all the $o_k(\Omega)$ simultaneously this way.
Edit: Here is the desired calculation for $AGL_1(\mathbb{F}_7)$ acting on $\mathbb{F}_7$. We need to compute the cycle decomposition of every affine map $x \mapsto ax + b, a \in \mathbb{F}_7^{\times}, b \in \mathbb{F}_7$.

*

*If $a = 1$, then $x \mapsto x + b$ is a translation, so it is a $7$-cycle if $b \neq 0$ and the identity if $b = 0$. The first case occurs $6$ times and the second case occurs once.

*If $a \neq 1$, then $x \mapsto ax + b$ has a unique fixed point $x = \frac{b}{1-a}$, and acts as scaling by $a$ centered at this fixed point. So, depending on whether $a$ has order $2, 3$, or $6$, it has a single fixed point and then either three $2$-cycles, two $3$-cycles, or one $6$-cycle. These cases occur $7, 2 \times 7, 2 \times 7$ times each respectively.

This means the generating function for $o_k(\mathbb{F}_7)$ in this case is
$$\begin{align*} \sum_k o_k(\mathbb{F}_7) t^k &= \frac{(1 + t)^7 + 6(1 + t^7) + 7(1 + t)(1 + t^2)^3 + 14(1 + t)(1 + t^3)^2 + 14(1 + t)(1 + t^6)}{42} \\
 &= \boxed{ 1 + t + t^2 + 2t^3 + 2t^4 + t^5 + t^6 + t^7 }.\end{align*}$$
Funnily enough the sequence is the same as for $GL_3(\mathbb{F}_2)$ acting on $\mathbb{P}^2(\mathbb{F}_2)$; I don't know if this is just a coincidence or if it has some interesting explanation. We get that the action on $k$-element subsets is transitive except for $k = 3, 4$ where it breaks up into two orbits. The polynomial above is symmetric since the action on $k$-element subsets is isomorphic to the action on their complements, and it's not hard to see that the action of the affine group is doubly transitive, so it remains to explain the $k = 3$ orbits.
These can be explained conceptually by thinking of the affine group as the subgroup of the projective group $PGL_2(\mathbb{F}_7)$ fixing $\infty$; it follows that the action on tuples of $3$ distinct points preserves the cross-ratio
$$(x, y, z, \infty) = \frac{z - x}{z - y}$$
which is also easy to see directly since the action of $x \mapsto ax + b$ multiplies both the numerator and the denominator by $a$. So the action on $3$-element subsets preserves the cross ratio up to the permutation action of $S_3$ acting on the $3$ points $\{ x, y, z \}$. The transposition $(xy)$ sends the cross-ratio $\lambda$ to $\frac{1}{\lambda}$, while the transposition $(yz)$ sends the cross-ratio $\lambda$ to $1 - \lambda$. Under the action of these two maps the set of possible cross-ratios breaks up into two orbits, namely $\{ 2, 4, 6 \}$ and $\{ 3, 5 \}$ ($1$ is not a possible value), and these correspond to the two orbits of $3$-element subsets.
