Let $K\_{n,n}$ be a complete bipartite graph with two parts $\{u\_1,u\_2,\ldots,u\_n\}$ and $\{v\_1,v\_2,\ldots,v\_n\}$, and let $K^-\_{n,n}$ be the graph derived from $K\_{n,n}$ by delete a perfect matching $\{u\_1v\_1,u\_2v\_2,\ldots,u\_nv\_n\}$.


Since $K^-\_{n,n}$ is now $(n-1)$-regular, it has $n-1$ disjoint perfect matchings. My question is whether the edges of $K^-\_{n,n}$ with $n\geq 4$ can be decomposed into $n-1$ disjoint perfect matchings in such a way that in each matching $M$, if $u\_iv\_j\in E(M)$ then $v\_iu\_j\not\in E(M)$.



Disjoint perfect matchings in complete bipartite graph
The answer is that this is possible for all $n>4$.


Your question is equivalent to asking whether there exists a unipotent Latin square $L$ of order $n$ with $L\_{ij}\ne L\_{ji}$ for $i\ne j$. The equivalence is obtained by using $L\_{ij}$ to record the index of the matching that contains the edge $u\_i v\_j$ (and putting $L\_{ii}=n$ for each $i$).


The existence of such a Latin square follows from a stronger property. It is a theorem (collectively due to the work of Kotzig, McLeish, Turgeon and others) that for all $n\notin\{2,4\}$ there exists a Latin square of order $n$ that has no intercalates (an intercalate is a $2\times2$ submatrix that is itself a Latin square). This property of Latin squares is called $N\_2$ in the literature.


If you permute the rows of any $N\_2$ Latin square you can make every symbol on the main diagonal equal to $n$. You then have what you need.


