Is every finite quantum group a quantum symmetry group? This post is basically a quantum extension of Is every finite group a group of “symmetries”?
Here finite quantum group means finite dimensional Hopf ${\rm C}^{\star}$-algebra.   
Frucht's theorem states that every finite group is the automorphism group of a finite graph. Wang defined here a notion of quantum automorphism group. The application to a finite space of $n$ elements is called "the quantum permutation group of $n$ symbols", and its quantum subgroups are called quantum permutation groups of degree $n$. Bichon introduced here the quantum automorphism groups of finite graphs, these are quantum permutation groups. See also this survey of Banica-Bichon-Collins. 
Kojima proved here that every finite group is realized as the full isometry group of some compact hyperbolic $3$-manifold. 
This book of Goswami-Bhowmick introduces the notion of quantum isometry group.  
General question: Is every finite quantum group a quantum symmetry group?   
Sub-question 1: Is a finite quantum permutation group a twisted finite group?  
Sub-question 2: Is every finite quantum group a quantum permutation group?
Answer: no, the smallest counter-example has dim. $24$ (see this paper of Banica-Bichon-Natale).  
Sub-question 3: Is every finite quantum permutation group of dimension $n$ a quantum permutation group of degree $n$?
Sub-question 4: Is every finite quantum permutation group a quantum automorphism group of a finite graph?
Sub-question 5: Is every finite quantum group $\mathbb{G}$ a quantum subgroup of the quantum automorphism group of a finite dimensional ${\rm C}^{\star}$-algebra $\mathcal{A}$? Ok for $\mathcal{A} = C(\mathbb{G})$?
Answer (Bhowmick): yes, it follows trivially from the definition, using Haar state.
Sub-question 6: Is every finite quantum group a quantum isometry group?
 A: The answer to sub-question 4 is no.
See here, The Frucht property in the quantum group setting.
To expand slightly, this paper gives four explicit finite quantum permutation groups which are not the quantum automorphism groups of a finite graph: the duals of $S_3$, $A_4$, and $A_5$, and the Kac--Paljutkin quantum group of order eight, denoted $G_0$.
If the dual of $S_3$ is the quantum automorphism group of a graph, then, taking abelianisations, the (classical) automorphism group of the graph must be $\mathbb{Z}_2$. The paper above shows that if the dual of $S_3$ acts on a graph then a (classical) abelian group larger than $\mathbb{Z}_2$ acts on the graph too. Therefore the dual of $S_3$ is not the quantum automorphism group of a graph.
In fact it is shown that if the dual of $S_3$ acts on a graph then the dual of a(n infinite) free product acts on the graph too.
A similar story holds for the duals of $A_4$ and $A_5$.
If $G_0$ is the quantum automorphism group of a graph, then the graph has classical automorphism group $\mathbb{Z}_2\times\mathbb{Z}_2$. The above shows that if $G_0$ acts on a graph then so does the dihedral group of order eight. Therefore $G_0$ is not the quantum automorphism group of a graph.
