Does this set sequence represent the Dedekind numbers? I got somehow interested in the set $X_n$ of mappings $\alpha:{\cal S}_n\rightarrow\{1,\dots,n\}$ such that for elements $p_1,\dots,p_n$ in $[n]$ we have $p_{\sigma(1)}\le\dots\le p_{\sigma(n)}$ and $p_{\sigma'(1)}\le\dots\le p_{\sigma'(n)}$ implies $p_{f(\sigma)}=p_{f(\sigma')}$.
Here $[n]$ denotes $\{1,\dots,n\}$ but in fact we can choose the $p_i$ to be of any poset with at least $n$ distinct comparable elements.  Note that the $p_i$ do not have to be distinct, otherwise the premise would be trivial.  Also note that $f(\sigma)$ does not have to be equal to $f(\sigma')$.
In other words, for any permutation $\sigma\in{\cal S}_n$ and adjacent transposition $\tau=(i,i+1)$ we want $\{f(\sigma),f(\sigma\circ\tau)\}\subseteq\{i,i+1\}$ or $f(\sigma)=f(\sigma\circ\tau)$.
After doing some coding, it seems that the sizes of $X_1,X_2,X_3,X_4$ are $1,4,18,166$ respectively.  On OESIS, the only sequence matching these numbers is a variation of the Dedekind numbers (it is always two below the 'original' Dedekind numbers, this can be thought of excluding two somehow trivial elements in every set of a corresponding set sequence, e.g. the constant boolean functions or the empty antichain and the antichain consisting of the empty set only).
So my question, is there a nice bijection between the $X_n$ and another set sequence known to represent those Dedekind numbers?
 A: I will construct a bijection between $X_n$ and the set $Y_n$ of reduced normal conjunctive forms in $n$ variables without negation (whose size is known to be the $n$th Dedeking number of the kind discussed above).  For $\sigma\in{\cal S}_n$ let $\sigma^*_i$ be the operator $\sigma\mapsto\sigma^{-1}(i)$.  Also let $\Sigma$ resp. $\Pi$ denote the signatures $\{\sigma^*_1,\dots,\sigma^*_n,\min,\max\}$ resp. $\{x_1,\dots,x_n,\land,\lor\}$ and $T(\Sigma)$ resp. $T(\Pi)$ the syntactic terms built over these signatures (without free variables).
Then I show that (i) $X_n$ is closed under $\min,\max$ and contains the constants $\sigma^*_i$, (ii) every $x\in X_n$ can be expressed by some term in $T(\Sigma)$ and (iii) the natural bijection between $T(\Sigma)$ and $T(\Pi)$ is compatible with term semantics, i.e. terms in $T(\Sigma)$ yielding the same mapping $\alpha:{\cal S}_n\to[n]$ are mapped to terms in $T(\Pi)$ yielding the same antichain ${\cal I}\subseteq{\cal P}([n])$ and vice versa.
(i) This is pretty much straight forward computation.
(ii) Fix $n$, I will show by induction on $k$ that if the range of $\alpha$ is contained in a $k$-element subset of $[n]$, then it is expressible over $T(\Sigma)$.  The constant $i$-function $c_i$ can be expressed as $$\min_{J\subseteq[n],|J|=i}\max_{j\in J}\sigma^*_j.$$  The hardest part to show is actually $k=2$.  First note that the range of $\alpha$ is always a complete intervall of natural numbers because if $\alpha(\sigma)=i$ and $\alpha(\sigma')=j$ then there are adjecent transpositions $\tau_1,\dots,\tau_m$ such that $\sigma'=\sigma\circ\tau_1\circ\dots\circ\tau_m$ so the sequence $\alpha(\sigma),\alpha(\sigma\circ\tau_1),\dots,\alpha(\sigma')$ connects $i$ with $j$ while jumping at most by one per step.  So in the particular case of $k=2$ the range of $\alpha$ is w.l.o.g. $\{i,i+1\}$.  If $\tau_1,\dots,\tau_m$ is a sequence of adjecent transpositions not containing $(i,i+1)$ then $\alpha(\sigma)$ must be equal to $\alpha(\sigma\circ\tau_1\circ\dots\circ\tau_m$).  From this follows that $\sigma([i])=\sigma'([i])$ implies $\alpha(\sigma)=\alpha(\sigma')$.  Let ${\cal I}$ be se set of $i$-element subsets of $[n]$ such that $\alpha(\sigma)=i$ if and only if $\sigma([i])$ is contained in ${\cal I}$, then $$\sigma=\min(c_{i+1},\min_{I\in{\cal I}}\max_{i\in I}\sigma^*_i.)$$  Now we can actually start with the induction step, w.l.o.g assume the image of $\alpha$ is contained in $[k]$.  Let $$\alpha_i(\sigma)=\begin{cases}1&\alpha(\sigma)=i\\i&\text{else}\end{cases}$$ for $2\le i\le k$ so the range of $\alpha_i$ is contained in a $k-1$-element set, then $$\sigma=\max_{2\le i\le k}\alpha_i(\sigma).$$
(iii) It is quite straight forward to check that $\min,\max$ adhere to the same reduction rules than $\land,\lor$ so we also have the same normal forms.  For some antichain ${\cal I}$ in $[n]$ denote with $\alpha_{\cal I}$ the map described by $$\min_{I\in{\cal I}}\max_{i\in I}\sigma^*_i,$$ it is to be shown that different ${\cal I}$'s produce different $\alpha_{\cal I}$'s.  So if ${\cal I}\neq{\cal J}$ there is w.l.o.g. some $I=\{i_1,\dots,i_k\}$ in ${\cal I}$ which is not superset of any $J$ in ${\cal J}$.  So pick some $\sigma$ with $\sigma(I)=[k]$, then $\alpha_{\cal I}(\sigma)=k$ but $\alpha_{\cal J}(\sigma)>k$.
