Given integers $1$ through $n$, let $S$ be set of ordering of integers that respect even alternating or reverse alternating permutations (https://en.wikipedia.org/wiki/Alternating_permutation) up to length $2n$ ($2,4\dots,n$ if $n$ is even, $2,4\dots,n-1$ if $n$ is odd), $\overline{S}$ be set of orderings that violate alternating and reverse alternating permutations up to length $2n$ .

$$3<4>2<5>1<6>3$$ $$4>2<5>1<6>3<4$$ $$2<3>1<4>2$$ $$4<5>3<6>3$$ are examples of alternating and reverse alternating permutations respectively of length up to $6$ and are in $S$.

$$3>2>1<6>5>4>3$$ $$4>2>1<5<6>3<4$$ $$4>2>1<3<4$$ $$6>4>3<5<6$$ are not alternating or reverse alternating permutations and are in $\overline{S}$.

$|S\cup\overline{S}|=n!$.

How many permutation maps $\sigma_1$ through $\sigma_K$ does one need so that for every $s\in S$ there is a $\sigma_i$ where $i\in\{1,\dots,K\}$ such that $\sigma_i(s)\in\overline{S}$?

Is $K>1$?

If $n=6$, $123456\rightarrow432165$, $123456\rightarrow456123$ both individually seem to convert every $s\in S$ to some element in $\overline S$.

**Update after Stanley's answer**
Answer provided below suggests $2$ permutations needed for every alternating and reverse alternating permutations cycle of given even length.

However in example with $n=6$ above, every even length alternating and reverse alternating permutations cycle of length at most $6$ cycles is covered ($4$ also is covered) with just $1$ permutation.

Stanley's answer provides a solution that needs at least $3$ permutations when $n=6$ ($\sigma_1$ is common for every even length cycle) or $n/2$ permutations asymptotically.

Do we need $K>1$ for this question or at least can $K\ll(\log n)^{1/a}$ with some fixed $a>1$ hold?