For this question, one can assume that the simplicial sets come from oriented simplicial complexes (you can assume the orientation induces an orientation on the geometric realization).
A morphism from a simplicial $X$ set into a groupoid $G$ is determined by the restriction of the morphism to the 2-skeleton of $X.$ In addition, a morphism $$f:\Delta^2\to G$$ is determined by $$f:\Lambda_i^2\to G,$$ for any $0\le i\le 2.$ This means that a morphism from the standard 2 simplex to $G$ is determined by the morphism on any horn of the standard 2 simplex, which is a one dimensional simplicial complex. Is something like this true if we replace $\Delta^2$ by any other simplicial complex? That is, given a simplicial complex $X,$ is there a one dimensional subsimplicial complex $Y\subset X$ such that $$\text{Hom}(X,G)=\text{Hom}(Y,G)?$$
In certain examples I can find such a $Y$ by deleting enough edges. My idea was to use the ordering to choose which edges to delete, but I'm not sure if this works in general.
There is a universal groupoid that every simplicial set maps to: its fundamental groupoid. To be precise, if $X$ is a simplicial set, consider the groupoid $\pi_{\leq 1}(X)$ whose objects are the vertices of $X$ and whose morphisms are homotopy classes of paths between them in the geometric realization. There is an obvious map $X\to\pi_{\leq 1}(X)$ and every map from $X$ to a groupoid factors uniquely through this map, the edges of $X$ generate $\pi_{\leq 1}(X)$ as a groupoid and homotopic paths formed by sequences of edges become equal under any map to a groupoid.
So, identifying $\operatorname{Hom}(X,G)$ with $\operatorname{Hom}(\pi_{\leq 1}(X),G)$, you are equivalently asking for a $1$-dimensional $Y\subseteq X$ such that $\pi_{\leq 1}(Y)=\pi_{\leq 1}(X)$. Since the fundamental group of a graph is always free, this is impossible unless the fundamental group of $X$ at each basepoint is free.
(I suspect this necessary condition is also sufficient. It would suffice to show that (assuming without loss of generality that $X$ is connected) you can find a $1$-dimensional subcomplex for which the inclusion induces an isomorphism on fundamental groups, since then you can extend that $1$-dimensional subcomplex to include every vertex without changing the fundamental group by adding edges one by one, as in the construction of a spanning tree.)