The number of posets with the same zeta polynomial.
The zeta polynomial $Z$ is the polynomial such that $Z(m)$ is the number of weakly increasing sequences $x_1\leq x_2\leq\dots\leq x_{m−1}$ of elements of the poset.
See section 3.12 of [1].
Since
$$ Z(q) = \sum_{k\geq 1} \binom{q-2}{k-1} c_k, $$
where $c_k$ is the number of chains of length $k$, this statistic is the same as the number of posets with the same chain polynomial.

The left-to-right maxima of a permutation as a Dyck path.
Let $(c_1, \dots, c_k)$ be the rise composition Mp00102rise composition of the path. Then the corresponding left-to-right maxima are $c_1, c_1+c_2, \dots, c_1+\dots+c_k$.
Restricted to 321-avoiding permutations, this is the inverse of Mp00119to 321-avoiding permutation (Krattenthaler), restricted to 312-avoiding permutations, this is the inverse of Mp00031to 312-avoiding permutation.

The cell poset of the parallelogram polyomino corresponding to the Dyck path.
Let $D$ be a Dyck path of semilength $n$. The parallelogram polyomino $\gamma(D)$ is defined as follows: let $\tilde D = d_0 d_1 \dots d_{2n+1}$ be the Dyck path obtained by prepending an up step and appending a down step to $D$. Then, the upper path of $\gamma(D)$ corresponds to the sequence of steps of $\tilde D$ with even indices, and the lower path of $\gamma(D)$ corresponds to the sequence of steps of $\tilde D$ with odd indices.
This map returns the cell poset of $\gamma(D)$. In this partial order, the cells of the polyomino are the elements and a cell covers those cells with which it shares an edge and which are closer to the origin.

Sends a permutation to its image under the Foata bijection.
The Foata bijection $\phi$ is a bijection on the set of words with no two equal letters. It can be defined by induction on the size of the word:
Given a word $w_1 w_2 ... w_n$, compute the image inductively by starting with $\phi(w_1) = w_1$.
At the $i$-th step, if $\phi(w_1 w_2 ... w_i) = v_1 v_2 ... v_i$, define $\phi(w_1 w_2 ... w_i w_{i+1})$ by placing $w_{i+1}$ on the end of the word $v_1 v_2 ... v_i$ and breaking the word up into blocks as follows.

• If $w_{i+1} \geq v_i$, place a vertical line to the right of each $v_k$ for which $w_{i+1} \geq v_k$.
• If $w_{i+1} < v_i$, place a vertical line to the right of each $v_k$ for which $w_{i+1} < v_k$.

In either case, place a vertical line at the start of the word as well. Now, within each block between vertical lines, cyclically shift the entries one place to the right.
To compute $\phi([1,4,2,5,3])$, the sequence of words is

• $1$
• $|1|4 \to 14$
• $|14|2 \to 412$
• $|4|1|2|5 \to 4125$
• $|4|125|3 \to 45123.$

In total, this gives $\phi([1,4,2,5,3]) = [4,5,1,2,3]$.
This bijection sends the major index (St000004The major index of a permutation.) to the number of inversions (St000018The number of inversions of a permutation.).