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suppose group G acts on group W,i.e.there is an injective hom from G to Aut(W). different injections give different actions.if the orbit spaces of two G actions on W are the same,on what ocassions, do we have the two actions are equivalent(i,e,the images of two injections from G to Aut(W) are conjugate in Aut(W)) Any comments on this question are welcome.

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In GL(2,11), there are two subgroups of order 55 (both non-abelian), which give the same orbit decompositions (on C_11 x C_11), but which are not conjugate in GL(2,11). – Steve D May 21 '10 at 16:37

No, ${\mathbb C}^\ast$ has a distinct action on ${\mathbb C}$ for each natural $n$, i.e. $x\cdot y = x^ny$ but the orbits structure is the same for all non-zero $n$.

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I don't really see why there should be general results. We can take W to be a vector space over a finite field, and then you are asking about equivalence of linear representations of G on W. The data you will give me is about the stabilisers H(w) of the w in W, and as I understand it there will be just this. But I think there are many examples where W will be very homogeneous (taking out 0). And there will be cases, for example, where there will be inequivalent linear representations of G that are related by outer automorphisms of G. If you tell me you can detect enough about the representations to determine their equivalence by some general method, without further hypotheses, I shall need convincing.

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