Notes on the ordered set $A^A$. Part IV. The dual ${}^A\text{\hspace{0.17em}⁣}A$ of $A^A$ for finite ordered sets

Let $A$ be a finite ordered set. Define the ordered set $A^A$ as the set of all maps from $A$ to $A$, ordered pointwise. Let ${}^{A}A$ be the dual of $A^A$. We prove results in the spirit of Parts~I–III, but now using both $A^A$ and ${}^{A}A$. For example, if $$({}^{{}^{{}^{{}^{A}A}A}A}A{)}^{A^{A^A}}$$ is isomorphic to $$({}^{{}^{{}^{{}^{B}B}B}B}B{)}^{B^{B^B}}$$ for finite ordered sets $A$ and $B$, then $A$ is isomorphic to $B$.