Let $e\_i \wedge e\_j \ (i < j)$ be a basis for the $\mathbb Z$-module $\wedge^2 \Gamma$, where $\Gamma = \mathbb Z^n$. 
Clearly $S\_n$ acts on the module $\wedge^2 \Gamma$ via 
$$\pi(e\_i \wedge e\_j) = e\_{\pi(i)} \wedge e\_{\pi(j)} \ \ \ \forall \pi \in S\_n.$$ By restriction this induces an action on the subset $\bar B = 
\{ \epsilon e\_i \wedge e\_j \ (i < j), \ \epsilon \in \{-1, 1\} \}$. 


Which (non-trivial) cyclic subgroups of $S\_n$ have maximal number of orbits in this action on $\bar B$.
The answer seems to be the subgroups generated by transpositions $\pi = (ij)$. But can there be other permutations $\pi$ that are not transpositions but with the same number of orbits?



action of symmetric group on the second exterior power
By the Lemma that is not Burnside's, the number of orbits is the average number of fixed points. An element fixes $\epsilon e\_i \wedge e\_j$ iff it fixes both $i,j$ (because if it swaps them it reverses the sign, and otherwise it won't even preserve the span $\mathbb{Z} e\_i \wedge e\_j$. It follows that $\sigma \in S\_n$ will have $2\binom{\#\mathrm{Fix}(\sigma)}{2}$ fixed points on $\bar{B}$ because we need to choose pairs $(i,j)$ in its fixed point set.


Now a transposition has the largest number of fixed points of any non-identity element. Accordingly let $G < S\_n$ be any non-trivial subgroup and let $C<S\_n$ be the subgroup generated by a transposition. Then we have


$ \# \bar{B}/G = \frac{1}{\# G} \sum\_{\sigma\in G} 2\binom{\#\mathrm{Fix}(\sigma)}{2} \leq \frac{1}{\# G} n(n-1) + \left(1-\frac{1}{\# G}\right)(n-2)(n-3) = (n-2)(n-3) + \frac{1}{\# G} (4n-6) \leq (n-2)(n-3) + \frac{1}{\# C} (4n-6) = \# \bar{B}/C \,.$


It follows that the number of orbits of $C$ is maximal, with equality iff $G$ is conjugate to $C$.


