Odd-length alternating-permutation equivalence of the six remaining patterns

Let $A_m(p)$ denote the set of alternating permutations of length $m$ that avoid the pattern $p$. Two patterns are equivalent for odd-length alternating permutations when their avoidance classes have equal cardinality for every odd length. Remaining odd-class conjecture. The patterns $2314$, $4132$, $2413$, $3142$, $1423$, and $3241$ are equivalent for odd-length alternating permutations; equivalently, for all $n\geq 0$ their avoidance classes have a common cardinality:

$$|A_{2n+1}(2314)|=|A_{2n+1}(4132)|=|A_{2n+1}(2413)|=|A_{2n+1}(3142)|=|A_{2n+1}(1423)|=|A_{2n+1}(3241)|.$$

The paper identifies this as the only other possible odd-length equivalence not ruled out by numerical data, and the conjecture remains open.