The question asks for the number of reduced three-line Latin rectangles with prescribed second row.  The best answer I know of is the following result due to Ira Gessel (<i>Combinatoire énumérative</i>,
Lecture Notes in Mathematics Volume 1234, 1986, pp.106&ndash;111):

><b>Theorem</b>.  The number of pairs $(\sigma,\tau)$ of permutations of
$\lbrace 1, 2, \ldots, n\rbrace$ such that $\sigma$, $\tau$, and $\sigma\tau^{-1}$
are fixed-point-free, $\sigma$ has $j$ cycles, and $\tau$ has $k$ cycles is the coefficient
of $\alpha^j\beta^k x^n\!/n!$ in
$$e^{2\alpha\beta x} \sum_{n=0}^\infty {\alpha^{\bar n}\beta^{\bar n}\over n!}
 {x^n \over (1+\alpha x)^{n+\beta} (1+\beta x)^{n+\alpha}(1+x)^{n+\alpha\beta}},$$
where $\alpha^{\bar n} := \alpha(\alpha+1)\cdots(\alpha+n-1)$.

Since we only care about the total number of $\tau$ for a given $\sigma$, we can set $\beta=1$ and just extract the coefficient of $\alpha^j x^n\!/n!$ and divide by the (unsigned) Stirling number of the first kind, $|s(n,j)|$.