Even-length alternating-permutation equivalence for eight patterns

Let $A_m(p)$ denote the set of alternating permutations of length $m$ that avoid the pattern $p$. Two patterns are equivalent for even-length alternating permutations when $|A_{2n}(p)|=|A_{2n}(q)|$ for every $n\geq 1$. Even-class conjecture. For all $n\geq 1$ and every $p\in\{1243,2134,1432,3214,2341,4123,3421,4312\}$,

$$|A_{2n}(p)|=|A_{2n}(1234)|=|A_{2n}(2143)|.$$

These are enumerative equivalences among length-four patterns; numerical data support the claim, while a proof, possibly via generating trees, remains open.