For even $$$n$$$, choosing the entire array in one operation will always turn it into a derangement. Therefore, the answer for even $$$n$$$ is just $$$n!$$$.

However, this does not work for odd $$$n$$$, because $$$\lceil \frac{n}{2} \rceil$$$ will fall into its corresponding index. Call this pivot value $$$v$$$. Notice that if $$$p_v = v$$$ and we can't move $$$v$$$ out of the $$$v^{th}$$$ index, then we can't turn $$$p$$$ into a derangement.

When does this happen? Well it must be the case that $$$p_v = v$$$ initially, all of the values to the left of $$$v$$$ are greater than $$$v$$$, and all of the values to the right of $$$v$$$ are less than $$$v$$$.

Why? Let's say one of the elements to the left of $$$v$$$ (at index $$$u$$$) is less than $$$v$$$. Then we can just apply the operation on $$$p_{u...v}$$$ and $$$v$$$ will be replaced with $$$u$$$ (or possibly a smaller element between the two indices). A similar argument can be made for the right side of $$$v$$$.

Now let's say the conditions are satisfied. We cannot move $$$p_v$$$ from $$$v$$$ because if we try to sort any subarray including index $$$v$$$, the order of the elements to the left of $$$v$$$ (or to the right of $$$v$$$) may change, but $$$v$$$ will stay in its position.

So now we have a condition where all $$$p$$$ satisfying it cannot be converted into some derangement. It turns out that all $$$p$$$ that don't satisfy the condition can always be turned into a derangement.

Convert the permutation into a simpler array $$$a$$$ where all elements less than $$$v$$$ are $$$0$$$'s, $$$v$$$ is $$$1$$$, and all elements greater than $$$v$$$ are $$$2$$$'s. The number of $$$0$$$'s is equal to the number of $$$2$$$'s. Now let's transform the operation into one that just swaps adjacent elements at indices $$$i, i + 1$$$ if and only if $$$a_i \lt a_{i + 1}$$$.

Notice that arrays of the form $$$[2, 2, \dots, 2, 0, 1, 0, 0, \dots 0]$$$ (form $$$1$$$) or $$$[2, 2, \dots, 2, 1, 2, 0, 0, \dots, 0]$$$ (form $$$2$$$) always represent derangements. For $$$n = 5$$$, these arrays would be $$$[2, 2, 0, 1, 0]$$$ and $$$[2, 1, 2, 0, 0]$$$, respectively.

The original $$$a$$$ won't be of the form $$$[2, 2, \dots, 2, 1, 0, 0, \dots, 0]$$$ (because such an array satisfies the condition). For the remaining arrays, we can either use the operation to push all of the $$$2$$$'s to the left and then position $$$1$$$ in its correct place to make form $$$1$$$, or we can use the operation to push all of the $$$0$$$'s to the right and then position $$$1$$$ in its correct place to make form $$$2$$$.

One of these procedures will certainly work because either the $$$1$$$ will come before a $$$2$$$ (meaning we can achieve form $$$2$$$) or the $$$1$$$ will come after a $$$0$$$ (meaning we can achieve form $$$1$$$).

So all permutations that don't satisfy the condition can be converted into some derangement. Therefore, the answer for odd $$$n$$$ is just $$$n! - (\lfloor \frac{n}{2} \rfloor!)^2$$$.