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Let G be a simple labeled graph with 12 vertices consisting of exactly two connected components which are path graphs on 6 vertices.

  o-o-o-o-o-o     o-o-o-o-o-o.

I think there are 4 automorphisms. Reverse one path, reverse the other path, or reverse both paths or do nothing (the identity mapping). If I try to count the labeleings I get binomial(12,6)*(6!/2)^2 and this is 12!/4. Everything checks. Right?

However, I am looking at data in Mathematica that says that this graph has 322560 automorphisms. This would imply 1485 labelings. Is this nonsense?

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I think there are $8$ automorphisms.

The $4$ that you found and $4$ extra, they consist of each of the previous $4$ and then swapping the two paths between each other.

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  • $\begingroup$ OK. That makes sense. My label count above is wrong. I need to divide binomial(12,6) by 2. This implies 8 automorphisms as you state. $\endgroup$ – Geoffrey Critzer Aug 9 '16 at 17:33

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