our definition of $S_n^{\varepsilon}$ coincides with the one of a quantum automorphism group of $\Gamma_{\varepsilon}$ given by Bichon [@BicQAG; @BicQAGZwei], see Proposition \[PropVglBB\]. In this sense, we may justify the definition $S_n^{\varepsilon}$ intrinsicly, i.e. as the quantum symmetry of some quantum space; exactly like we motivated our definition of $O_n^{\varepsilon}$ as the quantum symmetry of the ${\varepsilon}$-sphere. There is another definition of a quantum automorphism group of a graph given by Banica [@BanQAG]. We denote it by $S_n^{\Gamma_{\varepsilon}}$ in order to keep the notations used in this article consistent.

\[DefAut\] Given an undirected graph $\Gamma_{\varepsilon}$ with adjacency matrix ${\varepsilon}\in M_n(\{0,1\})$, its *quantum automorphism group* $S_n^{\Gamma_{\varepsilon}}$ is defined via: $$C(S_n^{\Gamma_{\varepsilon}}):=  C^*(u_{ij}\;|\; u_{ij}=u_{ij}^*=u_{ij}^2,\sum_ku_{ik}=\sum_ku_{kj}=1\;\forall i,j\textnormal{ and } u{\varepsilon}={\varepsilon}u)$$ More explicitely, $u{\varepsilon}={\varepsilon}u$ may be expressed as: $$\sum_k \delta_{{\varepsilon}_{kl}=1}u_{ik}=\sum_j \delta_{{\varepsilon}_{ij}=1}u_{jl}$$

\[LemRelSnEps\] Let $G\subset S_n^+$ be a quantum subgroup of $S_n^+$. The following relations are equivalent:

-   The partial relations of $R^{\varepsilon}$: $u_{ik}u_{jl}=u_{jk}u_{il}$ and $u_{ki}u_{lj}=u_{kj}u_{li}$ if ${\varepsilon}_{ij}=1$ and ${\varepsilon}_{kl}=0$.

-   The partial relations $(\mathring R^{\varepsilon}2)$ of $\mathring R^{\varepsilon}$: $\delta_{{\varepsilon}_{kl}=0}u_{ik}u_{jl}=\delta_{{\varepsilon}_{ij}=0}u_{ik}u_{jl}$.

-   The relations $u{\varepsilon}={\varepsilon}u$: $\sum_k \delta_{{\varepsilon}_{kl}=1}u_{ik}=\sum_j \delta_{{\varepsilon}_{ij}=1}u_{jl}$.

The equivalence of (i) and (ii) follows from the proof of Lemma \[LemREpsAeqRRing\]. We now prove that (ii) implies (iii). From (ii) we infer: $$\sum_{k,j}\delta_{{\varepsilon}_{kl}=1}\delta_{{\varepsilon}_{ij}=0}u_{ik}u_{jl}=0 \quad\textnormal{and}\quad
\sum_{k,j}\delta_{{\varepsilon}_{kl}=0}\delta_{{\varepsilon}_{ij}=1}u_{ik}u_{jl}=0$$ Thus, using $\sum_j 