In general, if $f : X \to Y$ is a morphism of schemes, then is it true that the diagonal morphism $\Delta : X \to X \times_Y X$ is the result of some base extension of $f$?
Specifically, I was onto the idea that, given $X \times_Y X$ is a scheme over $Y$, we might have that $X$ satisfies the universal fiber product properties with projections $id: X \to X$ and $\Delta : X \to X \times_Y X$. But working through the commutative diagrams, this doesn't appear to be the case unless I could assume e.g. $f$ is a monomorphism.
This is not possible in general. Let $Y = \operatorname{Spec} \mathbb Z$ be the final object.
If there were a cartesian diagram $$\require{AMScd} \begin{CD} X @>>> X\\ @V{\Delta}VV @V{f}VV \\ X \times_{Y} X @>>> \operatorname{Spec} \mathbb Z \end{CD}$$ we would have $$\operatorname{Hom}(-,X) = \operatorname{Hom}(-,X \times_{Y} X) \times \operatorname{Hom}(-,X) = \operatorname{Hom}(-,X) \times \operatorname{Hom}(-,X) \times \operatorname{Hom}(-,X)$$ , which is absurd for the most choices of $X$.