Assume $n \geq 2$. Let $\mathcal{M}_n$ denote the set of perfect matchings on $[2n] := \{1,\ldots,2n\}$, i.e., the set of partitions of $[2n]$ into pairs. For $M \in \mathcal{M}_n$, and $p = \{a,b\}$,$q = \{c,d\} \in M$ with $a < b$, $c < d$, and $a<c$, we say the pairs $p$ and $q$ are *aligned* if $a < b < c < d$. Denote the number of alignments of a matching $M$ by $\mathrm{al}(M) := |\{ \{p,q\} \subset M\colon p \textrm{ and } q \textrm{ are aligned}\}|$.

Is it possible to find a partition $O_n$ of $\mathcal{M}_n$ so that for every $\mathcal{O} \in O_n$ we have

- $|\mathcal{O}| = 3$;
- $\sum_{M \in \mathcal{O}} \mathrm{al}(M) = \binom{n}{2}$?

EDIT:

I can add the following context to the problem if it helps. Recall that we say $p$ and $q$ are *nesting* if $a < c < d < b$ and we say they are *crossing* if $a < c < b < d$. Let $\mathrm{ne}(M)$ denote the number of nestings of $M$, and $\mathrm{cr}(M)$ the number of crossings. The statement would follow easily if we had $\sum_{M \in \mathcal{M}_n} x_1^{\mathrm{ne}(M)}x_2^{\mathrm{cr}(M)}x_3^{\mathrm{al}(M)} = \sum_{M \in \mathcal{M}_n} x_{w(1)}^{\mathrm{ne}(M)}x_{w(2)}^{\mathrm{cr}(M)}x_{w(3)}^{\mathrm{al}(M)}$ for all $w \in S_3$, the symmetric group on three letters. However this is not the case: nestings, crossings, and alignments are not all equidistributed (although nestings and crossings are). Still, I expect a kind of $S_3$ symmetry with respect to nestings, crossings, and alignments in the form of homomesy (as the title of the question suggests). Recall from Propp and Roby that for a set of combinatorial objects $\mathcal{S}$ and an invertible map $\varphi\colon \mathcal{S} \to \mathcal{S}$ we say the statistic $f\colon \mathcal{S} \to K$ is *homomesic* with respect to the action of $\varphi$ on $\mathcal{S}$ if $\sum_{x \in \mathcal{O}}f(x)$ is constant among all $\varphi$-orbits $\mathcal{O}$.

Let $\mathrm{Aut}(\mathcal{M}_n)$ denote the group of invertible maps from $\mathcal{M}_n$ to $\mathcal{M}_n$. What I really want to find is some $\tau \in \mathrm{Aut}(\mathcal{M})$ with $\tau^3 = \mathrm{id}$ and such that $\mathrm{al}(\cdot)$ is homomesic with respect to the action of $\tau$ on $\mathcal{M}_n$. Ideally $\tau$ would also satisfy $(\tau \sigma)^2 = \mathrm{id}$, where $\sigma \in \mathrm{Aut}(\mathcal{M}_n)$ is the involution that swaps nestings and crossings defined in this paper of De Medicis and Viennot (see also this paper for a description of the algorithm in English). If this were possible, then defining $\Gamma := \langle \tau, \sigma \rangle \subset \mathrm{Aut}(\mathcal{M}_n)$, we would have $\Gamma \simeq S_3$ and all three of the statistics $\mathrm{ne}(\cdot)$, $\mathrm{cr}(\cdot)$, and $\mathrm{al}(\cdot)$ are homomesic with respect to the action of $\Gamma$ on $\mathcal{M}_n$. This would be interesting as it would be the first instance of non-cyclic (indeed, non-abelian) homomesy that I am aware of.