Theorem: Sets of Operations on Set of 3 Elements/Automorphism Group of A

:$\AA$ has $3$ elements.

Proof:
Recall the definition of (group) automorphism:
:$\phi$ is an automorphism on $\struct {S, \circ}$ {{iff}}:
::$\phi$ is a permutation of $S$
::$\phi$ is a homomorphism on $\struct {S, \circ}$: $\forall a, b \in S: \map \phi {a \circ b} = \map \phi a \circ \map \phi b$
Hence $\AA$ can be defined as the set of operations $\circ$ on $S$ such that every permutation on $S$ is an automorphism of $\struct {S, \circ}$.
The set $\map \Gamma S$ of all permutations on $S = \set {a, b, c}$ has $6$ elements.
From Identity Mapping is Group Automorphism, $I_S$ is always an automorphism on $\struct {S, \circ}$.
Hence it is not necessary to analyse the effect of $I_S$ on the various elements of $S$.
Let us denote each of the remaining permutations on $S$ as follows:
{{begin-eqn}}
{{eqn | l = p
      | o = :
      | r = \map p a = b, \map p b = c, \map p c = a
}}
{{eqn | l = q
      | o = :
      | r = \map q a = c, \map q b = a, \map q c = b
}}
{{eqn | l = r
      | o = :
      | r = \map r a = a, \map r b = c, \map r c = b
}}
{{eqn | l = s
      | o = :
      | r = \map s a = c, \map s b = b, \map s c = a
}}
{{eqn | l = t
      | o = :
      | r = \map t a = b, \map t b = a, \map t c = c
}}
{{end-eqn}}
So, let $\circ \in \AA$.
From Permutation of Set is Automorphism of Set under Left Operation and Permutation of Set is Automorphism of Set under Right Operation:
:${\to} \in \AA$
:${\gets} \in \AA$
where $\to$ and $\gets$ are the right operation and left operation respectively.
Next we note that from Structure of Cardinality 3+ where Every Permutation is Automorphism is Idempotent:
:$\forall x \in S: x \circ x = x$
Let us explore various options for $a \circ b$.
Let $a \circ b = a$.
Then by definition of the above mappings $p$, $q$, $r$, $s$ and $t$, and the definition of a homomorphism:
{{begin-eqn}}
{{eqn | l = \map p a \circ \map p b
      | r = \map p a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \circ c
      | r = b
      | c = 
}}
{{eqn | l = \map q a \circ \map q b
      | r = \map q a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \circ a
      | r = c
      | c = 
}}
{{eqn | l = \map r a \circ \map r b
      | r = \map r a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a \circ c
      | r = a
      | c = 
}}
{{eqn | l = \map s a \circ \map s b
      | r = \map s a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \circ b
      | r = c
      | c = 
}}
{{eqn | l = \map t a \circ \map t b
      | r = \map t a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \circ a
      | r = b
      | c = 
}}
{{end-eqn}}
This is none other than the left operation, which has already been counted.
{{qed|lemma}}
Let $a \circ b = b$.
Then by definition of the above mappings $p$, $q$, $r$, $s$ and $t$, and the definition of a homomorphism:
{{begin-eqn}}
{{eqn | l = \map p a \circ \map p b
      | r = \map p b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \circ c
      | r = c
      | c = 
}}
{{eqn | l = \map q a \circ \map q b
      | r = \map q b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \circ a
      | r = a
      | c = 
}}
{{eqn | l = \map r a \circ \map r b
      | r = \map r b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a \circ c
      | r = c
      | c = 
}}
{{eqn | l = \map s a \circ \map s b
      | r = \map s b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \circ b
      | r = b
      | c = 
}}
{{eqn | l = \map t a \circ \map t b
      | r = \map t b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \circ a
      | r = a
      | c = 
}}
{{end-eqn}}
This is none other than the right operation, which has already been counted.
{{qed|lemma}}
Let $a \circ b = c$.
Then by definition of the above mappings $p$, $q$, $r$, $s$ and $t$, and the definition of a homomorphism:
{{begin-eqn}}
{{eqn | l = \map p a \circ \map p b
      | r = \map p c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \circ c
      | r = a
      | c = 
}}
{{eqn | l = \map q a \circ \map q b
      | r = \map q c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \circ a
      | r = b
      | c = 
}}
{{eqn | l = \map r a \circ \map r b
      | r = \map r c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a \circ c
      | r = b
      | c = 
}}
{{eqn | l = \map s a \circ \map s b
      | r = \map s c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \circ b
      | r = a
      | c = 
}}
{{eqn | l = \map t a \circ \map t b
      | r = \map t c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \circ a
      | r = c
      | c = 
}}
{{end-eqn}}
Thus we have identified another operation on $S$ for which all permutations on $S$ are automorphisms of $\struct {S, \circ}$.
{{qed|lemma}}
We now note that once the product element $a \circ b$ has been selected to be either $a$, $b$ or $c$, the full structure of $\struct {S, \circ}$ is forced.
Hence there are no other operations $\circ$ on $S$ but these ones we have counted.
That is, there is a total of $3$ such operations.
{{qed}}