Edges of the complete graph on $2n$ vertices can be colored with $2n-1$ colors such that only edges of different colors intersect.
Can this always be done such that for every pair of different colors the set of edges of these two colors form a unique cycle?
Reformulation: Are there $2n-1$ fixed-point-free involutions in the symmetric group $S_{2n}$ such that the product of two (different) such involutions has always two cycles of length $n$?
(I suspect that the answer is well-known to specialists : This seems a fairly natural question.)
The answer is YES if $2n-1$ is a prime number. Indeed, dropping the condition that any maximal set of edges of two colors defines a hamiltonian cycle, there is an easy solution due to Soifer, see https://en.wikipedia.org/wiki/Edge_coloring which can be described as follows : Consider the set $S=\lbrace 0,1,\ldots,2n-2,\infty\rbrace$. For $a$ in $\lbrace 0,\ldots,2n-2\rbrace$ consider the fixed point-free involution $\sigma_a$ defined by $x\longmapsto 2a-x$ if $x\not\in \lbrace a,\infty\rbrace$ extended by $\sigma_a(a)=\infty$, $\sigma_a(\infty)=a$. A symmetry argument shows that it is enough to prove the claim for all pairs of colors $\lbrace \sigma_0,\sigma_a\rbrace$ for finite non-zero $a$. The composition of these two involutions is then essentially addition (or subtraction) of $2a$ modulo $2n-1$. We get therefore Hamiltonian cycles for all choices of $a$ if $2n-1$ is prime. We have however a problem if $a$ divides $2n-1$.
