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We are taught that, in general:

A type of objects that has nontrivial automorphisms cannot have a fine moduli space.

The proof generally goes along the lines of:

Take an object $X$ with a non-trivial automorphism $\phi$. Form a twisted family on the circle $S^1$ by taking $[0,1]\times X$ and identifying $(0,x)$ with $(1,\phi(x))$ (in a category other than $\mathbf{Top}$ this might need some work, or be impossible). Then the resulting family is non-trivial (but see below).

Suppose this family corresponded to some morphism $f\colon S^1\to M$, where $M$ is a fine moduli space for the problem. But the fibres are all isomorphic, so the map $f$ must be constant. But that means that the pullback along $f$ is trivial, and so cannot be equal to our family, which is a contradiction.

There are a few problems which might occur, outlined at this section of nLab. But in many cases, the existence of automorphisms does prevent the existence of a moduli space.

But take this example: a family of oriented topological circles over a space $X$ is a fibre bundle $E\xrightarrow{p}X$ with fibre $S^1$, and a continuous choice of orientation for the fibres; i.e., a group action of $S^1$ on $E$ which acts fibre-wise. In other words, it is a principal $S^1$-bundle.

But it is known that the automorphism classes of principal $S^1$-bundles over a space correspond to its second homology group. So any principal $S^1$-bundle over the circle (say) is trivial, and there are no examples arising in the manner outlined above. If we only looked at spaces with zero second homology, we might conclude that a moduli space for oriented topological circles is given by the point.

But there certainly are examples of non-trivial principal $S^1$-bundles (since there are spaces with non-zero second homology). The simplest example is the Hopf bundle over $S^2$. So there cannot be a fine moduli space for oriented topological circles. But this example doesn't arise in any way from the existence of non-trivial automorphisms of the oriented circle; of course, the circle has got non-trivial automorphisms, but they vary continuously among all circles; there are certainly no special circles which have extra automorphisms, as there are in the elliptic curve case.

Is there a more general obstruction to the existence of a moduli space for a moduli problem? You can restrict attention to moduli problems over $\mathbf{Top}$. It looks as if the original non-trivial-automorphism examples correspond to the first homology/fundamental group, but the oriented circle example corresponds to second homology, and it's not hard to construct similar examples for $n$-th homology using Eilenberg-Maclane spaces.

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    $\begingroup$ The Hom-functor $F$ between two affine schemes of finite type over a noetherian ring $k$ (e.g., the Hom-functor from the affine line to itself is the functor of "polynomials of unspecified degree") is generally not representable but this has nothing to do with automorphisms. Such $F$ satisfies the functorial criterion to be locally of finite presentation, so if representable then must be so by a $k$-scheme $M$ locally of finite type, yet $F$ generally has infinite-dimensional tangent space at geometric points. This obstruction is not repaired by working with stacks. $\endgroup$
    – user74230
    Apr 15, 2015 at 13:33
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    $\begingroup$ X-posted to math.SE: math.stackexchange.com/questions/1235631/… $\endgroup$ Apr 15, 2015 at 13:34

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