Notes on the ordered set $A^A$. Part IV. The dual ${}^{A}\!A$ of $A^A$ for finite ordered sets

Abstract: 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 [ \Bigl({}^{{}^{ {}^{ {}^{A}A}A}A}A\Bigr)^{A^{A^{A}}} ] is isomorphic to [ \Bigl({}^{ {}^{ {}^{ {}^{B}B}B}B}B\Bigr)^{B^{B^{B}}} ] for finite ordered sets $A$ and $B$, then $A$ is isomorphic to $B$.