Killing the automorphisms to make a functor representable

Question

I was told (quickly) by my professor about the following yoga:

Say we want to define a functor $F:\operatorname{Sch}\to \operatorname{Set}$ which parametrizes some class of objects. If those objects have nontrivial automorphisms, then we cannot expect $F$ to be representable. There are two solutions to this issues:

1. Sheafify the functor $F$ in order to obtain an fppf sheaf
2. Rigidify the objects we want to parametrize, in order to kill their automorphisms

Now, at the present moment I don't have the time to go through the underlying theory. So

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Could you give me some intuition on the above yoga?

Why are the automorphisms an obstruction to the representability of the functor?

What does it mean, in simple terms, for a functor to be a fppf sheaf?

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I have a motivating example for this question:

Let $X$ and $T$ be schemes over $S$; the relative Picard functor $Pic_{X/S}:Sch_S \to Ab$ is defined by $$ T \mapsto \left\{ \text{Rigidified line bundles}\text{ on } X_T/T \right\} $$ and not as $$ T \mapsto \left\{ \text{Line bundles}\text{ on } X_T/T \right\} $$ The main difference between line bundles and rigidified ones is that the latter don't have any nontrivial automorphisms.

Under reasonable hypothesis the first functor turns out to be representable, while the first one doesn't.

Answer 1

Regarding #2, here is a related question. The idea is rather simple: the existence of nontrivial automorphisms implies existence of nontrivial isotrivial families (i.e. with all fibres isomorphic/equivalent). To see why this messes up representability, suppose we are given a nontrivial isotrivial family $\mathcal X\to B$. There exists a unique map $B\to P$, where $P$ supposedly represents whatever moduli functor we're interested in, such that $\mathcal X\to B$ is the pullback over this map of the universal family $\mathcal U\to P$. The map $B\to P$ is clearly constant, by the fact that the fibres are isomorphic. But this is impossible, since the pullback of $\mathcal U\to P$ over a constant map is the trivial family.

Hopefully someone else can say something about #1.