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I want to know to what extent is the group action determined by its fixed point data and orbit data, i.e. if $G$ acts on $M$ in two ways with the same fixed point set and orbit space, on what occasions it is true that the two ways are equivalent (i.e.the two actions are conjugate in $Homeo(M)$)?

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What category is M in? –  Qiaochu Yuan Apr 29 '10 at 15:03
    
I think we can consider the problem in topological category or differential category –  sara Apr 29 '10 at 22:54
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2 Answers

First, you might want to identify actions that are not conjugated in $\mathrm{Homeo}(M)$. For example, let $\mathbb{Z}\times\mathbb{Z}$ act on $M$ in the following two ways: the first factor acts by a map $f:M\to M$ and the second acts trivially, or the converse. Then the actions are the same only up to an automorphism of the group.

Second, it is not clear to me what you mean by "the same orbit space". Rational rotations of the circle all have homeomorphic orbit spaces, but of course they are not conjugate in any way.

I guess that even in favorable cases, you need the (conjugacy class of the) stabilizer of each orbit to identify the action.

Last, a remark that is not directly linked to your question, but that I like to advertise: there exist (infinite families) of analytic group actions on analytic manifolds that are topologically but not $C^1$ conjugate.

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In http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&s1=143095&vfpref=html&r=41&mx-pid=338129 Quillen worked some equivalent categories:
Proposition1 :
If $\mathcal{C}$ is any small category, in which every morphism is invertible, and $G:= C$ (as groupoid) then (for example) the category of $G$-actions is equivalent to the category of covering spaces of the geometric realization of $\mathcal{C}$.

The proof of the proposition is, except for some covering-spaces-arguments, not too difficult. Furthermore, maybe the occurring functors might lead you on the right track.

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