This question follows up a previous question, [Intersecting group orbits](https://mathoverflow.net/questions/90826/intersecting-group-orbits).


Suppose a group $G$ acts transitively on a set $X$ of $n$ elements, where $n$ is even, and consider the induced action on the set $\binom{X}{n/2}$ of subsets of $X$ of size $n/2$.


Question 1: Can there be precisely two orbits? More generally, if there is more than one orbit, how many orbits must there be?


Question 2: Suppose that no set is in the orbit of its complement. How many orbits must there be? Can there be precisely two orbits?



Intersecting group orbits, version 2
There's a cute example with $n=6$. Start with a regular icosahedron; it has 12 vertices and 20 triangular faces. Identify antipodal points. Now you have 6 points and 10 triangles. Let $H$ be the group of those permutations of the vertices that send triangles to triangles; this $H$ has order 60. $H$ is an index-2 subgroup of a group $G$ of permutations of the six vertices such that each of the permutations in $G-H$ sends your 10 triangles to exactly the 10 3-eleemt sets of vertices that aren't among your triangles. So $G$ has exactly 2 orbits on the set of all 20 3-element sets of vertices: the 10 triangles and the 10 other 3-element sets.


It turns out that, in this situation, a 3-element set is one of the 10 triangles iff its complement isn't. So the 3-element sets in one orbit are exactly the complements of the 3-element sets in the other orbit. This is, except for a trivial example with $n=2$, the only example that does what you asked and has this additional complementation property.


As a bonus: It is well known that 6 is the only $n$ for which the symmetric group $S\_n$ has outer automorphisms. Under these outer automorphisms of $S\_6$, the subgroup $G$ corresponds to the stabilizer of a point. 


