A 1-factorisation of a graph is perfect if the union of any two of its
1-factors is a Hamiltonian cycle.  Let $n=p^2$ for an odd prime $p$.
We construct a family of $(p-1)/2$ non-isomorphic perfect
1-factorisations of $K_{n,n}$.  Equivalently, we construct
pan-Hamiltonian Latin squares of order $n$. A Latin square is
pan-Hamiltonian if the permutation defined by any row relative to any
other row is a single cycle.