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

:$\BB$ has $3^3 - 3$ 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 $\struct {S, \circ}$: $\forall a, b \in S: \map \phi {a \circ b} = \map \phi a \circ \map \phi b$
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 $S$.
Let us denote each of the remaining elements of $\set {I_S, \tuple {a, b, c}, \tuple {a, c, b} }$ 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
}}
{{end-eqn}}
We select various product elements $x \circ y \in S$ and determine how $p$ and $q$ constrain other product elements as follows:
Then by definition of the above mappings $p$ and $q$, and the definition of a homomorphism, we obtain as follows:
;$(1): \quad a \circ a$
{{Begin-table}}|-
| align="left"  |
{{begin-eqn}}
{{eqn | l = a \circ a
      | r = a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map p a \circ \map p a
      | r = \map p a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \circ b
      | r = b
      | c = 
}}
{{eqn | l = \map q a \circ \map q a
      | r = \map q a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \circ c
      | r = c
      | c = 
}}
{{end-eqn}}
| align="left"   |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & a &   &   \\
b &   & b &   \\
c &   &   & c \\
\end {array}$
|-
| align="left"  |
{{begin-eqn}}
{{eqn | l = a \circ a
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map p a \circ \map p a
      | r = \map p b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \circ b
      | r = c
      | c = 
}}
{{eqn | l = \map q a \circ \map q a
      | r = \map q b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \circ c
      | r = a
      | c = 
}}
{{end-eqn}}
| align="left"   |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & b &   &   \\
b &   & c &   \\
c &   &   & a \\
\end {array}$
|-
| align="left"  |
{{begin-eqn}}
{{eqn | l = a \circ a
      | r = c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map p a \circ \map p a
      | r = \map p c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \circ b
      | r = a
      | c = 
}}
{{eqn | l = \map q a \circ \map q a
      | r = \map q c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \circ c
      | r = b
      | c = 
}}
{{end-eqn}}
| align="left"   |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a & c &   &   \\
b &   & a &   \\
c &   &   & b \\
\end {array}$
{{End-table}}
So selecting $a \circ a$ fixes $b \circ b$ and $c \circ c$.
;$(2): \quad a \circ b$
{{Begin-table}}|-
| align="left"  |
{{begin-eqn}}
{{eqn | l = a \circ b
      | r = a
      | c = 
}}
{{eqn | ll= \leadsto
      | 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 = 
}}
{{end-eqn}}
| align="left"   |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   & a &   \\
b &   &   & b \\
c & c &   &   \\
\end {array}$
|-
| align="left"  |
{{begin-eqn}}
{{eqn | l = a \circ b
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | 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 = 
}}
{{end-eqn}}
| align="left"   |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   & b &   \\
b &   &   & c \\
c & a &   &   \\
\end {array}$
|-
| align="left"  |
{{begin-eqn}}
{{eqn | l = a \circ b
      | r = c
      | c = 
}}
{{eqn | ll= \leadsto
      | 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 = 
}}
{{end-eqn}}
| align="left"   |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   & c &   \\
b &   &   & a \\
c & b &   &   \\
\end {array}$
{{End-table}}
So selecting $a \circ b$ fixes $b \circ c$ and $c \circ a$.
;$(3): \quad a \circ c$
{{Begin-table}}|-
| align="left"  |
{{begin-eqn}}
{{eqn | l = a \circ c
      | r = a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map p a \circ \map p c
      | r = \map p a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \circ a
      | r = b
      | c = 
}}
{{eqn | l = \map q a \circ \map q c
      | r = \map q a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \circ b
      | r = c
      | c = 
}}
{{end-eqn}}
| align="left"   |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   &   & a \\
b & b &   &   \\
c &   & c &   \\
\end {array}$
|-
| align="left"  |
{{begin-eqn}}
{{eqn | l = a \circ c
      | r = b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map p a \circ \map p c
      | r = \map p b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \circ a
      | r = c
      | c = 
}}
{{eqn | l = \map q a \circ \map q c
      | r = \map q b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \circ b
      | r = a
      | c = 
}}
{{end-eqn}}
| align="left"   |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   &   & b \\
b & c &   &   \\
c &   & a &   \\
\end {array}$
|-
| align="left"  |
{{begin-eqn}}
{{eqn | l = a \circ c
      | r = c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map p a \circ \map p c
      | r = \map p c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b \circ a
      | r = a
      | c = 
}}
{{eqn | l = \map q a \circ \map q c
      | r = \map q c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = c \circ b
      | r = b
      | c = 
}}
{{end-eqn}}
| align="left"   |
:$\begin {array} {c|ccc}
\circ & a & b & c \\
\hline
a &   &   & c \\
b & a &   &   \\
c &   & b &   \\
\end {array}$
{{End-table}}
So selecting $a \circ c$ fixes $b \circ a$ and $c \circ b$.
There are $3$ elements $x$ of $S$ with which $a \circ x$ can be made.
For each of these $3$, there are $3$ different elements $y$ such that $a \circ x = y$.
These collectively fix all the possible values of $b \circ x$ and $c \circ x$.
Hence they exhaust all possible operations $\circ$ on $S$ for which $p$ and $q$ are automorphisms.
Thus there are independently:
:$3$ different options for $a \circ a$
:$3$ different options for $a \circ b$
:$3$ different options for $a \circ c$
and therefore $3 \times 3 \times 3 = 3^3$ operations $\circ$ on $S$ for which $p$ and $q$ are automorphisms.
Note that from Automorphism Group of $\AA$, $3$ of these are also such that the group of automorphisms of $\struct {S, \circ}$ is the symmetric group on $S$.
So these are excluded from our count.
The result follows.
{{qed}}