and only if the automorphism group $\mathrm{Aut}(G)$ is unimodular. An important special case of this is the following. Let $K$ be as above with a proper, admissible (i.e. the stabilizer of a simplex is the intersection of the stabilizers of its vertices) and cocompact action of a discrete normal subgroup $\Gamma\unlhd \mathrm{Aut}(K)=:G$. Let ${\mathcal F}$ be a fundamental domain for the $\Gamma$ action and $\{x_1,x_2,...\}$ a complete orbit section for $G$ in $V(K)$. Remark that $G/\Gamma$ is also discrete since ${\mathcal F}$ is finite. Subsequently, we have the following formula for the [*rrsc*]{}: $$\begin{aligned}
    \mu_{K}&:=\frac{1}{c}\sum_i \frac{\delta_{[K,x_i]}}{|\mathrm{Stab}_G(x_i)|}=\frac{1}{c}\sum_i \sum_{\substack{g\in G\\ gx\in V({\mathcal F})}}\frac{\delta_{[K,gx_i]}|\mathrm{Stab}_{G/\Gamma}(\Gamma x_i)|}{|\mathrm{Stab}_G(x_i)||G/\Gamma|}\\
    &=\frac{1}{c}\sum_{y\in V({\mathcal F})}\frac{\delta_{[K,y]}}{|G/\Gamma| |\mathrm{Stab}_{\Gamma}(y)|}\end{aligned}$$ And in a similar way we obtain that $$c=\sum_{y\in V({\mathcal F})} \frac{|G/\Gamma|}{|\mathrm{Stab}_\Gamma(y)|},$$ hence, with $c(\Gamma)=\sum_{y\in V({\mathcal F})} (|\mathrm{Stab}_\Gamma(y)|)^{-1}$, we deduce that $$\begin{aligned}
    \mu_K:=\frac{1}{c(\Gamma)}\sum _{y\in V({\mathcal F})}\frac{\delta_{[K,y]}}{|\mathrm{Stab}_{\Gamma}(y)|}.\end{aligned}$$ This shows that $\mu_K$ does not depend on the group or the choice of the fundamental domain. Note that if the $\Gamma$-action is free, $\mu_K$ has the following appearance: $$\mu_K= \sum _{y\in V({\mathcal F})}\frac{\delta_{[K,y]}}{|V({\mathcal F})|}.$$

3.  Probability measure preserving equivalence relations $\mathcal{R}\subset X\times X$ on a standard Borel space $(X,\nu)$ also give rise to random rooted simplicial complexes. Let $\Sigma$ be an $\mathcal{R}$-simplicial complex as introduced by Gaboriau 