Why the existence of automorphism of varieties makes a functor not being a fine moduli space?

Question

Let $F: (Sch) \to (Sets)$ be a functor sends schemes to sets (for example, $F$ sends a scheme $S$ to families of K3 surfaces over $S$ with some fixed polarization). Then it is known that because of the existence of non trivial automorphism (in above example, it is because the non trivial automorphism of K3 surfaces), this functor cannot be represented by a scheme (i.e. we do not have $F \cong Hom(-,M)$).

My question is why the existence of automorphism makes a functor not being representable?

Answer 1

Here's one concrete example that I like. Consider the functor $\mathcal{M}_{1,1} : (Sch)^{op} \to Set$. The $B$-valued points of this functor are elliptic curves over $B$. Now let us now consider two elliptic curves

$$E_1: y^2 = x^3 + x + 1, \hspace{5mm} E_2: 3y^2 = x^3 + x + 1.$$

It is not hard to show these are non-isomorphic over the rationals, but become isomorphic when we pass to the extension $\Bbb{Q}(\sqrt{3})$.

Conclusion: The presence of automorphisms implies that the functor $\mathcal{M}_{1,1}$ is not a sheaf in the fpqc topology, and hence cannot be representable by a scheme.