I looked and couldn't find the reference, so I might be wrong about this faithfulness result. Since the question doesn't depend on it, I'll edit it out...

@GeoffroyHorel by loop spectrum of $X$ I mean $\Omega^\infty \Sigma^\infty(X)$. You're right and I was wrong that Mandel's result is about cochains with prime coefficients. Thank you also for linking Allen's article, which I didn't know about. But both of these are results about reconstruction of a space from Eilenberg-McLane cohomology theories. What I was referring to (and seem to remember hearing somewhere) is that the assignment $X\mapsto S(X)$ is fully faithful without any Frobenius, with the partial inverse $E\mapsto Hom(E, S)$ (hom space in the category of $E_\infty$ algebras).

Thank you. I had not read Ayoub's papers and they seem to be very close to what I want. And Jon -- I've seen your papers before, but didn't realize their motivic context - thanks. I also found the answer to this question to be helpful: mathoverflow.net/questions/254341/…

Another vague answer is that the E2 operad is canonically a Z/2-equivariant operad via complex conjugation, which a very natural structure to consider for "real homotopy theory" reasons (for example the configuration spaces of n points in the plane are complex points of varieties defined over Q). The (non-equivariant) E1 operad is the fixed point sub-operad. Combining this conjucation equivariance structure with the natural cyclic Z/2 action on E2 should give a Z/2 equivariant E1 operad as some sort of twisted fixed points.

For example, a Riemannian manifold with a choice of nonvanishing vector field $\xi$ does not "reduce structure" to O(n-1), as there can be multiple non-isomorphic triples $(V, g, \xi)$ with $g$ a metric and $\xi\in V$: this is because the length of $\xi$ (w.r.t. g) is an isomorphism invariant. So a Riemann manifold with vector field is not classified by any $G$-bundle (it is not a "symmetry structure"). You can eliminate this extra parameter by requiring $|\xi| = 1$ or only fixing $\xi$ up to positive scalar.

@Student No -- and the issue is precisely this concept of "ceasing to be a symmetry structure". You need a piece of functorial information that has what is known as "no moduli".

@crystalline Thanks! That's exactly what I was hoping for -- will read the paper. If you can turn that into an answer, I'll approve it.

@crystalline I meant of course "quasi-smooth". "Complicated" is in the eye of the beholder, but the definitions I knew of seemed so to me. Thank you for the Schurg, Toen, Vezzosi reference -- that seems very close to what I need (even closer, though probably non-existent, is Lurie's assertion in HAG that stable maps are a representable DG moduli problem.

What you wrote is usually formulated as the fact that a DG module over a smooth, proper DG algebra is perfect if and only if it is perfect over the ground field. That is indeed true, though not all interesting algebras are smooth and perfect.