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10. $S$ is the non-unital semilattice defined by the following table (a semilattice is a commutative semigroup in which every element is idempotent):

11. $S$ is the unital semilattice (with identity/bottom element $x$) defined by the following table:

10. $S$ is the unital semilattice (with identity/bottom element $x$) defined by the following table (a semilattice is a commutative semigroup in which every element is idempotent):

11. $S$ is the non-unital semilattice defined by the following table:

As a complement to my previous non-answer, let me show that the OP's conjecture holds for any order-$3$ semigroup $S$ that is not a group (equivalently, that is not a cyclic group of order $3$).

Up to the canonical anti-isomorphism between $S$ and its opposite semigroup, there are $17$ such semigroups (excluding the cyclic group of order $3$), of which only $11$ are commutative; see the Wikipedia article on order-$3$ semigroups here. Below, I'll list these $11$ commutative semigroups one by one and prove that they all satisfy the OP's conjecture. It may be a tedious task, but is going to help with another reduction.

All semigroups will be defined on the set $\{x, y, z\}$ through their Cayley tables. Since $\lfloor (3-1)/2 \rfloor = 1$, we only need to find, for each semigroup on the list, a permutation $(a,b,c)$ of the triple $(x,y,z)$ such that $ac = bc$.

10. $S$ is the non-unital semilattice defined by the following table (a semilattice is a commutative semigroup in which every element is idempotent):

11. $S$ is the unital semilattice (with identity/bottom element $x$) defined by the following table (a semilattice is a commutative semigroup in which every element is idempotent):

As a complement to my previous non-answer, let me show that the OP's conjecture holds for any order-$3$ commutative semigroup $S$ that is not a group (equivalently, that is not a cyclic group of order $3$).

Up to the canonical anti-isomorphism between a semigroup and its opposite, there are $17$ semigroups of order $3$ (excluding the cyclic group of order $3$), of which only $11$ are commutative; see the Wikipedia article on order-$3$ semigroups here. Below, I'll list these $11$ commutative semigroups one by one and prove that they all satisfy the OP's conjecture. It may be a tedious task, but is going to help with another reduction.

All semigroups will be defined on the set $\{x, y, z\}$ through their Cayley tables. Since $\lfloor (3-1)/2 \rfloor = 1$, we need to find, for each semigroup on the list, a permutation $(a,b,c)$ of the triple $(x,y,z)$ such that $ac = bc$.

10. $S$ is the non-unital semilattice defined by the following table (a semilattice is a commutative semigroup in which every element is idempotent):

11. $S$ is the unital semilattice (with identity/bottom element $x$) defined by the following table:

As a complement to my previous non-answer, let me show that the OP's conjecture holds for any order-$3$ semigroup $S$ that is not a group.

Up to the canonical anti-isomorphism between $S$ and its opposite semigroup, there are $17$ such semigroups (excluding the cyclic group of order $3$), of which only $11$ are commutative; see the Wikipedia article on order-$3$ semigroups here. Below, I'll list these $11$ commutative semigorups one by one and prove that they all satisfy the OP's conjecture. It may be a tediousv task, but is going to help with another reduction.

All semigroups will be defined on the set $\{x, y, z\}$ through their Cayley tables. Since $\lfloor (3-1)/2 \rfloor = 1$, we need to find, for each semigroup on the list, a permutation $(a,b,c)$ of the ordered triple $(x,y,z)$ such that $ac = bc$.

As a complement to my previous non-answer, let me show that the OP's conjecture holds for any order-$3$ semigroup $S$ that is not a group (equivalently, that is not a cyclic group of order $3$).

Up to the canonical anti-isomorphism between $S$ and its opposite semigroup, there are $17$ such semigroups (excluding the cyclic group of order $3$), of which only $11$ are commutative; see the Wikipedia article on order-$3$ semigroups here. Below, I'll list these $11$ commutative semigroups one by one and prove that they all satisfy the OP's conjecture. It may be a tedious task, but is going to help with another reduction.

All semigroups will be defined on the set $\{x, y, z\}$ through their Cayley tables. Since $\lfloor (3-1)/2 \rfloor = 1$, we only need to find, for each semigroup on the list, a permutation $(a,b,c)$ of the triple $(x,y,z)$ such that $ac = bc$.