**Proof.**
Assume (2). The category $\mathcal{I}_\mathcal {S}$ has objects $(x, \alpha )$ where $x \in \mathcal{S}$, say with $p(x) = U$, and $\alpha : x \to x$ is a morphism in $\mathcal{S}_ U$. Hence if $\mathcal{I}_\mathcal {S} \to \mathcal{S}$ is an equivalence over $\mathcal{C}$ then every pair of objects $(x, \alpha )$, $(x, \alpha ')$ are isomorphic in the fibre category of $\mathcal{I}_\mathcal {S}$ over $U$. Looking at the definition of morphisms in $\mathcal{I}_\mathcal {S}$ we conclude that $\alpha $, $\alpha '$ are conjugate in the group of automorphisms of $x$. Hence taking $\alpha ' = \text{id}_ x$ we conclude that every automorphism of $x$ is equal to the identity. Since $\mathcal{S} \to \mathcal{C}$ is fibred in groupoids this implies that $\mathcal{S} \to \mathcal{C}$ is fibred in setoids. We omit the proof of (1) $\Rightarrow $ (2).
$\square$

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