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First, let me point out that I think ARupinski is right and $\rm{Aut}(\Gamma)$ is simply $S_{c-1}$. (This could be relevant.)

Next, note that there are at least 11 orbits on $V(\Gamma)\times V(\Gamma)$.

There is an obvious partition into 9 parts (coming from the 3 orbits on $V(\Gamma)$) each belonging to a different orbit, but 2 of these split into two orbits, a diagonal part and a non-diagonal part.

For example $(2,2)$ is in a different orbit than $(2,3)$.

In fact, there are even more orbits than this. For example, because the part corresponding to $s\times s$ splits even further : $({1,2},{2,3})$ ({1,2},{2,3}) is not in the same orbit as $({1,2},{3,4})$. ({1,2},{3,4}). The first is an ordered pair of vertices having a neighbour in common, while the second is a pair of vertices with no neighbour in common.

EDIT: As Dima Pasechnik explained, two of the other parts also split further in two, for a total of 14 orbits.

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First, let me point out that I think ARupinski is right and $\rm{Aut}(\Gamma)$ is simply $S_{c-1}$. (This could be relevant.)

Next, note that there are at least 11 orbits on $V(\Gamma)\times V(\Gamma)$.

There is an obvious partition into 9 parts (coming from the 3 orbits on $V(\Gamma)$) each belonging to a different orbit, but 2 of these split into two orbits, a diagonal part and a non-diagonal part.

For example $(2,2)$ is in a different orbit than $(2,3)$.

EDIT: Also

In fact, I think ARupinski is right and there are even more orbits than this, because the part corresponding to $\Aut(\Gamma)$ s\times s$splits even further :$({1,2},{2,3})$is simply not in the same orbit as$S_{c-1}$.({1,2},{3,4})$. The first is an ordered pair of vertices having a neighbour in common, while the second is a pair of vertices with no neighbour in common.

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There are 11 orbits on $V(\Gamma)\times V(\Gamma)$.

There is an obvious partition into 9 parts (coming from the 3 orbits on $V(\Gamma)$) each belonging to a different orbit, but 2 of these split into two, a diagonal part and a non-diagonal part.

For example $(2,2)$ is in a different orbit than $(2,3)$.

EDIT: Also, I think ARupinski is right and $\Aut(\Gamma)$ is simply $S_{c-1}$.