Existence of a possible counterexample in automaton semigroups In an attempt to resolve a question posed by Cain in his paper on Automaton Semigroups (open problem 6.12), I would like to know if there exists a finite semigroup $S$ satisfying the following properties:


*

*$S$ is self-dual (anti-isomorphic to itself)

*$S\neq S^2$

*$S^2$ is a band

*$S$ has a faithful left-regular representation (i.e.all rows in the Cayley Table of $S$ are distinct)


I'm having no luck constructing one (or proving that one doesn't exist) so any help/suggestions would be greatly appreciated!
 A: I believe I have an example but you should check the details of whether it works.  Maybe GAP can be used.  
Let $X=\{1,2,3,4,5\}$ and $X'=\{1',2',3',4',5'\}$.  Let $a,b\colon X\to X$ be given by $$a=\begin{pmatrix}  1 & 2 & 3& 4 &5\\ 2& 3& 3 &4& 5 \end{pmatrix}\qquad b=\begin{pmatrix}  1 & 2 & 3& 4 &5\\ 4& 5& 4 &4& 5 \end{pmatrix}.$$  Let $T=\langle a,b\rangle$ where we view $a,b$ as functions acting on the right of $X$.   Then one checks $a\neq a^2=a^3$, $b^2=b$, $ba=b$ and I suppose these are defining relations.Anyway $T=\{a,a^2,b,ab,a^2b\}$.   In particular $T^2$ is a band and $T^2\neq T$.  Crucial is that $ab\neq a^2b$.
Let $T'=\langle a',b'\rangle$ be the dual semigroup obtained by reversing the multiplication of $T$.  So $a'b'=b'$ and still $a'\neq (a')^2=(a')^3$, $(b')^2=b'$. Also $b'a'\neq b'(a')^2$. We view $T'$ as functions acting on the left of $X'$ in the obvious way (replace $i$ by $i'$ for each $i$ in $X$ to get $a',b'$ from $a$ and $b$). Let $\overline {T}$ be the subsemigroup of $T'\times T$ generated by $(a',a)$ and $(b',b)$.  It has $11$ elements and satisfies $\overline{T}^2$ is a band, $\overline{T}^2\neq \overline{T}$ and $\overline{T}$ is self-dual via the obvious involution.  $\overline{T}$ almost acts faithfully on the left of itself except that elements of the form $(b'a',t)$ and $(b'(a')^2,t)$ act the same for any $t\in \{b,ab,a^2b\}$.
To remedy that let $R=X'\times X$ with the rectangular band multiplication $(i,j)(k,l)=(i,l)$.  Then $S=\overline T\cup R$ is a semigroup using the products in $\overline{T}$ and $R$ already defined and by putting $(u,v)(i,j)=(u(i),j)$ and $(i,j)(u,v)=(i,jv)$ for $u\in T', v\in T$, $i\in X'$ and $j\in X$. So $R$ is the minimal ideal of  $S$.  Note $S^2$ is a band, $S^2\neq S$ and still $S$ is self-dual (using the obvious involution on $X'\times X$ and the involution on $\overline{T}$).
Because $T'$ acts faithfully on the left of $X'$ we can now distinguish $(b'a',t)$ and $(b'(a')^2,t)$ (with $t$ as above) by the action on the left of $R$.  Clearly if $i\neq k$ then $(i,j)$ and $(k,l)$ do not act the same on the left of $R$.  On the other hand if $\{j,k\}\neq \{2,3\}$, then $$(i,j)(a',a)=(i,ja)\neq (i,ka)=(i,k)(a',a).$$  On the other hand $$(i,2)(b',b)=(i,5)\neq (i,4)=(i,3)(b',b).$$ Thus the action of $S$ on the left of itself is faithful.  Note that $S$ has $36$ elements. 
I hope this is correct and helps.
A: Not an answer, but a question based on a previous attempt of finding a solution, for which Benjamin Steinberg pointed out a foolish mistake in the comments below. Is there an "enlarge-and-shrink recipe" to extend a semigroup $\mathbb A = (A, \cdot)$ for which Conditions 1, 3 and 4 hold true to a larger semigroup $(S, \cdot)$ for which Conditions 1, 3 and 4 continue to be true, but in addition $S^2 \ne S$? If the answer is yes, then the problem is solved in the positive: Start with your preferred self-dual band $\mathbb A$, unitize it by adjoining an identity only if $\mathbb A$ is not already unital (in such a way that the unitization is still a self-dual band, but we have a gain in the process, since now the outcome is a semigroup whose regular representations are both faithful, regardless as to whether or not this was already the case with $\mathbb A$), and finally use the enlarge-and-shrink recipe to conclude.
