  A $[k,n,1]$-graph is a $k$-partite graph with parts of order $n$ such that
the bipartite graph induced by any pair of parts is a matching. An independent
transversal in such a graph is an independent set that intersects each part in
a single vertex. A factor of independent transversals is a set of $n$
pairwise-disjoint independent transversals. Let $f(k)$ be the smallest integer
$n_0$ such that every $[k,n,1]$-graph has a factor of independent transversals
assuming $n \ge n_0$. Several known conjectures imply that for $k \ge 2$,
$f(k)=k$ if $k$ is even and $f(k)=k+1$ if $k$ is odd. While a simple greedy
algorithm based on iterating Hall's Theorem shows that $f(k) \le 2k-2$, no
better bound is known and in fact, there are instances showing that the bound
$2k-2$ is tight for the greedy algorithm. Here we significantly improve upon
the greedy algorithm bound and prove that $f(k) \le 1.78k$ for all $k$
sufficiently large, answering a question of MacKeigan.
