Assume that for some odd $n$ we have a complete graph $K_n$. Is it possible to always find Eluerian trail on $K_n$ such that if we have visited vertices $axb$ then we never before nor after visit $ayb$ nor $bya$ for some $y$. Basically if I split my Eulerian trail into paths of length two I want each two points to be endpoints of a single such path.

For prime $n$ I can find such Eulerian trail by just labelling vertices 1,2,...,n and then doing hamiltonian cycles starting from 1 where first cycle goes 1,2,...,n,1 ; second 1,3,5,7,...,n-1,1 and so on. But I am not sure how to do it for composite $n$ since in this case this isn't an Eulerian trail. Does someone have a reference or an idea how this would work?

Assume that for some odd $n$ we have a complete graph $K_n$. Is it possible to always find Eluerian trail on $K_n$ such that if we have visited vertices $axb$ then we never before nor after visit $ayb$ nor $bya$ for some $y$. Basically if I split my Eulerian trail into paths of length two I want each two points to be endpoints of a single such path.

For prime $n$ I can find such Eulerian trail by just labelling vertices 1,2,...,n and then doing hamiltonian cycles starting from 1 where first cycle goes 1,2,...,n,1 ; second 1,3,5,7,...,n-1,1 and so on. But I am not sure how to do it for composite $n$ since in this case this isn't an Eulerian trail. Does someone have a reference or an idea how this would work?

Edit: Thanks to the comment of Ilya Bogdanov below I see that my attempt was very obviously incorrect. The question of weather such an eulerian trail exists is still interesting to me,

Assume that for some odd $n$ we have a complete graph $K_n$. Is it possible to always find Eluerian trail on $K_n$ such that if we have visited vertices $axb$ then we never before nor after visit $ayb$ nor $bya$ for some $y$. Basically if I split my Eulerian trail into paths of length two I want each two points to be endpoints of a single such path.

For small $n$ I can find such Eulerian trail by just labelling vertices 1,2,...,n and then doing hamiltonian cycles starting from 1 where first cycle goes 1,2,...,n,1 ; second 1,3,5,7,...,n-1,1 and so on. But I am not sure if this works for some large $n$. Does anyone see why this approach would clearly fail or work?

Assume that for some odd $n$ we have a complete graph $K_n$. Is it possible to always find Eluerian trail on $K_n$ such that if we have visited vertices $axb$ then we never before nor after visit $ayb$ nor $bya$ for some $y$. Basically if I split my Eulerian trail into paths of length two I want each two points to be endpoints of a single such path.

For prime $n$ I can find such Eulerian trail by just labelling vertices 1,2,...,n and then doing hamiltonian cycles starting from 1 where first cycle goes 1,2,...,n,1 ; second 1,3,5,7,...,n-1,1 and so on. But I am not sure how to do it for composite $n$ since in this case this isn't an Eulerian trail. Does someone have a reference or an idea how this would work?