Sets of Operations on Set of 3 Elements/Automorphism Group of A/Commutative Operations

Theorem
Let $S = \set {a, b, c}$ be a set with $3$ elements.

Let $\AA$ be the set of all operations $\circ$ on $S$ such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$, that is, $\map \Gamma S$.

Then:
 * Exactly $1$ of the operations of $\AA$ is commutative.

Proof
Recall from Automorphism Group of $\AA$ the elements of $\AA$, expressed in Cayley table form:


 * $\begin {array} {c|ccc}

\to & a & b & c \\ \hline a & a & b & c \\ b & a & b & c \\ c & a & b & c \\ \end {array} \qquad \begin {array} {c|ccc} \gets & a & b & c \\ \hline a & a & a & a \\ b & b & b & b \\ c & c & c & c \\ \end {array} \qquad \begin {array} {c|ccc} \circ & a & b & c \\ \hline a & a & c & b \\ b & c & b & a \\ c & b & a & c \\ \end {array}$

From Cayley Table for Commutative Operation is Symmetrical about Main Diagonal, it is apparent by inspection that of the above:
 * $\begin {array} {c|ccc}

\circ & a & b & c \\ \hline a & a & c & b \\ b & c & b & a \\ c & b & a & c \\ \end {array}$ is the only Cayley table of a commutative operation.