, 2006

Let us take $n = 4k$. First we observe that such $\sigma$ cannot have a fixed point; for if $\sigma(j) = j$ for some $j$, then $\sigma^{-1}(j) = j$ and hence $n + 1 = \sigma(j) + \sigma^{-1}(j) = 2j$, which is impossible because $n$ is divisible by $4$.

Suppose $\sigma(a) = b$, where $a \neq b$. Then $\sigma^{-1}(a) = n + 1 - b$; $\sigma(b) = n + 1 - a$ and $\sigma(n + 1 - a) = n + 1 - b$. Thus we obtain a $4$-cycle $(a, b, n+1-a, n+1-b)$. It is easy to check that all these are distinct. Thus any such permutation $\sigma$ is a product of $n/4$ disjoint $4$-cycles of the above type. Note that $b$ and $n + 1 - b$ may be interchanged to get another admissible cycle $(a, n+1-b, n+1-a, b)$.

Thus we need to split the set $\{1, 2, 3, \ldots, n\}$ into $n/2$ pairs $\{j, n+1-j\}$, $1 \leq j \leq n/2$ and partition these $n/2$ pairs into $n/4$ pairs of pairs. Each such pair of pairs gives $2$ admissible $4$-cycles. The first splitting can be done in $\frac{1}{k!} \binom{2k}{2} \binom{2k-2}{2} \cdots \binom{2}{2} = \frac{(2k)!}{2^k k!}$ ways and there are $2^k$ further ways of getting $4$-cycles. Hence the desired number is
$$\frac{(2k)!}{2^k k!} = \frac{(2k)!}{k!} = 2 \cdot 6 \cdot 10 \cdots (n-2).$$