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Let me see if I understand your example correctly: you are fixing $X$ and $Y$, families of curves over $S$, and now you are considering the functor which maps an $S$-scheme $T$ to the set of $T$-isomorphisms $f^*X \to f^*Y$ (where $f$ is the map from $T$ to $S$).

If I have things straight, then this functor shouldn't be so bad to think about, because it is actually representable, by an Isom scheme. In other words, there is an $S$-scheme $Isom_S(X,Y)$ whose $T$-valued points, for any $f:T \to S$, are precisely the $T$-isomorphisms from $f^*X$ to $f^*Y$. (One can construct the Isom scheme by looking inside a certain well-chosen Hilbert scheme.)

One way to think about this geometrically is as follows: one can imagine that two curves over $k$ (a field) are isomorphic precisely when certain invariants coincide (e.g. for elliptic curves, the $j$-invariant). (Of course this is a simplification, and the whole point of the theory of moduli spaces/schemes/stacks is to make it precise, but it is a helpful intuition.) Now if we have a family $X$ over $S$, these invariants vary over $S$ to give a collections of functions on $S$ (e.g. a function $j$ in the genus $1$ case), and similarly with $Y$. Now $X$ and $Y$ will have isomorphic fibres precisely at those points where the invariants coincide, so if we look at the subscheme $Z$ of $S$ defined by the coincidence of the invariants, we expect that $f^*X$ and $f^*Y$ will be isomorphic precisely if the map $f$ factors through $Z$. Thus $Z$ is a rough approximation to the Isom scheme.

It is not precisely the Isom scheme, because curves sometimes have non-trivial automorphisms, and so even if we know that $X_s$ and $Y_s$ are isomorphic for some $s \in S$, they may be isomorphic in more than one way. So actually the Isom scheme will be some kind of (possibly ramified) finite cover of $Z$.

If one thinks about it carefully, one sees that the "arrow functor

Of course, if one pursues this line of intuition much more seriously, one will recover the notions of moduli stack, coarse moduli space, and so on.

The families $X$ and $Y$ over $S$ correspond to a map $\phi:S \to {\mathcal M}_g \times {\mathcal M}_g$. The stack which maps a $T$-scheme to $Isom_T(f^*X, f^*Y)$ can then seen to be the fibre product of the map $\phi$ and the diagonal $\Delta:{\mathcal M}_g \to {\mathcal M}_g \times {\mathcal M}_g$.

In the particular case of ${\mathcal M}_g$ the fact that this fibre product is representable is part of the condition that ${\mathcal M}_g$ be an algebraic stack.

But in general, the construction you describe is the construction of a fibre product with the diagonal. This might help with the geometric picture, and make the relationship to Mike's answer clearer. (For the latter:note that the path space into $X$ has a natural projection to $X\times X$ (take the two endpoints), and the loop space is the fibre product of the path space with the diagonal $X\to X\times X$.)

2 added 1018 characters in body

If one thinks about it carefully, one sees that the "arrow functorrecover the notions of moduli stack, coarse moduli space, and so on.

The families $X$ and $Y$ over $S$ correspond to a map $\phi:S \to {\mathcal M}_g\times {\mathcal M}_g$. The stack which maps a $T$-scheme to $Isom_T(f^*X,f^*Y)$ can then seen to be the fibre product of the map $\phi$ and the diagonal$\Delta:{\mathcal M}_g \to {\mathcal M}_g \times {\mathcal M}_g$.

In the particular case of ${\mathcal M}_g$ the fact that this fibre product is representable is part of the condition that ${\mathcal M}_g$ be an algebraic stack.

But in general, the construction you describe is the construction of a fibre productwith the diagonal. This might help with the geometric picture, and make the relationship to Mike's answer clearer. (For the latter:note that the path space into $X$ has a naturalprojection to $X\times X$ (take the two endpoints), and the loop space is the fibre productof the path space with the diagonal $X\to X\times X$.)

1

Let me see if I understand your example correctly: you are fixing $X$ and $Y$, families of curves over $S$, and now you are considering the functor which maps an $S$-scheme $T$ to the set of $T$-isomorphisms $f^*X \to f^*Y$ (where $f$ is the map from $T$ to $S$).

If I have things straight, then this functor shouldn't be so bad to think about, because it is actually representable, by an Isom scheme. In other words, there is an $S$-scheme $Isom_S(X,Y)$ whose $T$-valued points, for any $f:T \to S$, are precisely the $T$-isomorphisms from $f^*X$ to $f^*Y$. (One can construct the Isom scheme by looking inside a certain well-chosen Hilbert scheme.)

One way to think about this geometrically is as follows: one can imagine that two curves over $k$ (a field) are isomorphic precisely when certain invariants coincide (e.g. for elliptic curves, the $j$-invariant). (Of course this is a simplification, and the whole point of the theory of moduli spaces/schemes/stacks is to make it precise, but it is a helpful intuition.) Now if we have a family $X$ over $S$, these invariants vary over $S$ to give a collections of functions on $S$ (e.g. a function $j$ in the genus $1$ case), and similarly with $Y$. Now $X$ and $Y$ will have isomorphic fibres precisely at those points where the invariants coincide, so if we look at the subscheme $Z$ of $S$ defined by the coincidence of the invariants, we expect that $f^*X$ and $f^*Y$ will be isomorphic precisely if the map $f$ factors through $Z$. Thus $Z$ is a rough approximation to the Isom scheme.

It is not precisely the Isom scheme, because curves sometimes have non-trivial automorphisms, and so even if we know that $X_s$ and $Y_s$ are isomorphic for some $s \in S$, they may be isomorphic in more than one way. So actually the Isom scheme will be some kind of (possibly ramified) finite cover of $Z$.

Of course, if one pursues this line of intuition much more seriously, one will recover the notions of moduli stack, coarse moduli space, and so on.