Proof.
Part (1) follows Categories, Lemma 4.35.15 as (\mathit{Sch}/S)_{fppf} has fibre products.
Let a, b : x \to y be morphisms of \mathcal{X}. Set U = p(x) and V = p(y). The category of schemes has equalizers hence we can let W \to U be the equalizer of p(a) and p(b). Denote c : z \to x a morphism of \mathcal{X} lying over W \to U. The equalizer of a and b, if it exists, is the equalizer of a \circ c and b \circ c. Thus we may assume that p(a) = p(b) = f : U \to V. As \mathcal{X} is fibred in groupoids, there exists a unique automorphism i : x \to x in the fibre category of \mathcal{X} over U such that a \circ i = b. Again the equalizer of a and b is the equalizer of \text{id}_ x and i. Recall that the \mathit{Isom}_\mathcal {X}(x) is the presheaf on (\mathit{Sch}/U)_{fppf} which to T/U associates the set of automorphisms of x|_ T in the fibre category of \mathcal{X} over T, see Stacks, Definition 8.2.2. If \mathit{Isom}_\mathcal {X}(x) is representable by an algebraic space G \to U, then we see that \text{id}_ x and i define morphisms e, i : U \to G over U. Set M = U \times _{e, G, i} U, which by Morphisms of Spaces, Lemma 67.4.7 is a scheme. Then it is clear that x|_ M \to x is the equalizer of the maps \text{id}_ x and i in \mathcal{X}. This proves (2).
If \mathcal{X} = [U/R] for some groupoid in algebraic spaces (U, R, s, t, c) over S, then the hypothesis of (2) holds by Bootstrap, Lemma 80.11.5. If \mathcal{X} is an algebraic stack, then we can choose a presentation [U/R] \cong \mathcal{X} by Algebraic Stacks, Lemma 94.16.2.
\square
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