**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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