Can you say more about how to choose f and g? For (f(x), g(x,y)) to be area-preserving, g has to be linear in y, which feels like too strong a condition. I should say what I know about Gromov's nonsqueezing theorem in the question. What I know isn't much, but it contradicts your intuition about being able to get anything as the image.

Are you really saying that there are square-zero extensions of the sphere spectrum that are unramified? This isn't possible with plain rings.

OK not "sections of the conormal bundle" but "functions on the total space of the conormal bundle." A vector field on X determines a function on the conormal bundle to Y by contraction, and operates on the vector space of such functions by multiplication.

They aren't simply connected! Even for SL_2. The centralizer of [[0,0],[1,0]] has one component of the form [[1,0][,1]] and one of the form [[-1,0][,-1]].

I'm sorry, I had in mind that the equivariant homotopy type of a G-space just was its diagram of fixed-point spaces. This equivariant Whitehead theorem is almost very satisfying, but I feel like there is something hiding in the definition of G-CW complex. Is every G-manifold (with some tameness conditions if you like, like the manifold and group action are real analytic) strongly G-homotopy equivalent to a G-CW complex? That would seal the deal.

I think you are recovering the homotopy type of X, not X itself. In fact you can recover, tautologically, the equivariant homotopy type of G->Aut(X) out of the diagram O(G)->Spaces. Is there some other natural way of saying "equivariant homotopy type" so that this tautology becomes an interesting theorem?

I'm worried about things like the skyscraper D-module on A^1 supported away from the origin. I think that the Fourier transform of this does not have regular singularities, so how can it commute with Riemann-Hilbert? But if I understand you, you're saying that I should be looking at G_m-equivariant sheaves on all of these things, and then there will be a nice comparison theorem. When you put it like that...