The $n$-dimensional analogue of the octahedron is the complement of a perfect matching of its vertex set. (Every vertex is joined to every other vertex except its antipode.) If you take a Hamilton cycle in the $n$-dimensional octahedron then you can think of that as a labelling of the $2n$-cycle, and the non-edges of the $n$-octahedron give a perfect matching of the complement of this graph.
This paragraph was added in answer to Nicodean's comment below. Label the vertices of the $n$-dimensional octahedron $1^+,1^-,...,n^+,n^-$, where the vertices $i^+$ and $i^-$ are antipodes. A labelled Hamilton cycle in the octahedral graph is then exactly a $2n$-cycle made from the $2n$ symbols $i^+,i^-$ with the property that the symbols $i^+$ and $i^-$ are never adjacent. Giving a perfect matching on the complement of a $2n$-cycle is the same as giving a matching of the elements of a $2n$-cycle so that adjacent elements are never matched. To do this, I choose the symbols making up my matching to be the symbols $1^+,1^-,...,n^+,n^-$, where the symbols $i^+$ and $i^-$ are pairs that are matched. The condition that adjacent symbols are never matched is exactly the same condition that I had on the $2n$-cycles that describe the Hamilton cycles in the $n$-octahedral graph.
One reason why this seems hard is that the matching sort of stays the same, while the $2n$-cycle whose complement it lives in is the thing that changes when you change the Hamilton cycle.