Counting some special derangements

A derangement of a list of $n$ distinct entries is a permutation of that list such that no corresponding entries match. It is well-known that the number of such derangements is the nearest integer to $n!/e$ where $e$ is the base of natural logarithms.

Let's say that a permutation of a list is a $\mu$-derangement if both it and its reverse ordering are derangements. Equivalently, iff the permutation is a derangement both of the original list and of the reverse of the original list.

How many $\mu$-derangements of the list $[1,2,..,n]$ are there? Is there an exact formula? A good approximation or bound?

There are no $\mu$-derangements for $n \lt 4$, except in the trivial case of an empty list. I got counts for $n \le 10$ below by enumerating possibilities with a bit of Prolog code:

 n | # of µ-derangements 
---+----------------------
 4 |           4
 5 |          16
 6 |          80
 7 |         672
 8 |        4752
 9 |       48768
10 |      440192

The OEIS has this sequence as A003471, with a recurrence relation that suggests some separation into even and odd terms might simplify things.

For $n=2k$, this is equivalent to What's the General Expression For Probability of a Failed Gift Exchange Draw, with a slot and its reverse partner forming a couple – the answer for this case is

$$\int_0^\infty\left(x^2-4x+2\right)^k\mathrm e^{-x}\mathrm dx\;,$$

and I derived in the answer to the other question that this goes as $n!/\mathrm e^2$ for large $n$.

For $n=2k+1$, there's a single slot without a partner which leads to a factor $-L_1(x)=x-1$, so the answer for this case is

$$\int_0^\infty\left(x^2-4x+2\right)^k(x-1)\mathrm e^{-x}\mathrm dx\;.$$

The asymptotic analysis remains essentially unchanged, so this also goes as $n!/\mathrm e^2$ for large $n$.