I am trying to apply the main theorem of [this paper](http://www.tandfonline.com/doi/abs/10.1080/03081080500092232) to a certain kind of graph and keep getting confused. The theorem uses $rank(Aut\Gamma)$ which is defined as "the number of $Aut \Gamma$ orbits on the set $V(\Gamma) \times V(\Gamma)$". ($\Gamma$ is a graph).


Now, my graph $\Gamma$ is built in this way: take a clique of $c$ vertices, labelled $\{1,2,\ldots,c\}$ and add $s=\binom{c-1}{2}$ additional vertices, each of which is connected to a different pair of two vertices from $\{2,\ldots,c\}$.



> 
> Question: What is $rank(Aut\Gamma)$?
> 
> 
> 


My answer is 9 because the automorphism group is (apparently) $S\_{c-1} \times S\_{s}$ and there are 3 orbits for it. However, when I plug 9 into the theorem I get a contradiction with the rest of it (which involves objects I have a better grasp of so I am pretty sure I got the rest right).


Therefore, I suspect that my answer to the above question is wrong and I am in dire need of some enlightenment.


EDIT: Let's assume $c \geq 4$ to rule out sporadic cases.



Confused about orbits
The automorphism group of this graph is $S\_{c-1}$. Note that the vertex 1 in your clique cannot be moved anywhere (look at the degrees). On the other hand, a permutation of the remaining vertices in {2,...,c} induces a permutation on these $s$ vertices $P$. 
The 14 orbitals (aka orbits on $V\times V$) are as follows:


1. {(1,1)}
2. {(x,x) | x in {2..c}}
3. {(p,p) | p in P}
4. {(x,y) | x,y in {2..c}, x not equal to y}
5. {(p,q) | p,q in P, the corresponding pairs of elements of {2..c} do not intersect}
6. {(p,q) | p,q in P, the corresponding pairs of elements of {2..c} intersect in one element}
7. {(1,x) | x in {2..c}}
8. {(x,1) | x in {2..c}}
9. {(1,p) | p in P}
10. {(p,1) | p in P}
11. {(x,p) | x in {2..c}, p in P, x in p}
12. {(p,x) | x in {2..c}, p in P, x in p}
13. {(x,p) | x in {2..c}, p in P, x not in p}
14. {(p,x) | x in {2..c}, p in P, x not in p}


