Hey Math Overflow!
Say we have a (simple) graph Gamma, $\Gamma$, and G is the automorphism group of GammaG=Aut($\Gamma$) .
Is it true (in general) that 2 induced subgraphs of Gamma, $\Gamma$, say Gamma1 $\Gamma_1$ and Gamma2, $\Gamma_2$, are isomorphic iff they are in the same orbit of the action of G?
I suspect that the answer is 'no'.
First, I think one side is true and is trivial: If they are in the same orbit then there is an automorphism that, restricted to the vertices of Gamma1 $\Gamma_1$ and Gamma2 $\Gamma_2$ is an isomorphism.
However, I don't know, that given an isomorphism between Gamma1 $\Gamma_1$ and Gamma2 $\Gamma_2$ if we can extend it to an automorphism on Gamma.$\Gamma$.
Am I correct so far?
Also, given a graph, how do I go about to show that for this specific graph this argument is true (while not being true in general)? I suspect that it has some connection to the cycle index of the action of G on V.
I know that Z(G,1+x) $Z(G,1+x) = 1+x+2x^2+4x^3+5x^4+5x^5+4x^6+...+x^91+x+2x^2+4x^3+5x^4+5x^5+4x^6+...+x^9$
(the graph in question is L_2(3))$L_2(3)$)
Thanks in advance!
Edit1: Proper notations.