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.

$\begingroup$ In GL(2,11), there are two subgroups of order 55 (both nonabelian), which give the same orbit decompositions (on C_11 x C_11), but which are not conjugate in GL(2,11). $\endgroup$ – 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 nonzero $n$.

$\begingroup$ Although really it's not a yes or no question, but a "when" question, which is more interesting (though too general). $\endgroup$ – LSpice Sep 4 at 0:52

$\begingroup$ How sure are you? I cannot remember 10 years back but my answer predates the last edition of the question. I think I answered the first edition... $\endgroup$ – Bugs Bunny Sep 7 at 18:37

$\begingroup$ I thought the same at first, but the only change between the two editions was the addition of the word 'space' in the title: mathoverflow.net/posts/25463/revisions . $\endgroup$ – LSpice Sep 7 at 22:44
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.