Given a fusion category $\mathcal C$, the *Grothendieck Ring* $K\_0(\mathcal C)$ is the $\mathbb Z$-based ring whose basis elements correspond to isomorphism classes of simple objects and whose multiplication is given by


$$
X\times Y = \sum\_Z N\_{XY}^Z Z, N\_{XY}^Z=\vert \mathcal C(X\otimes Y,Z)\vert
$$


Two fusion categories $\mathcal C$ and $\mathcal D$ are said to be Grothendieck equivalent if $K\_0(\mathcal C)\cong K\_0(\mathcal D)$.


Given $\mathcal C$, the *adjoint subcategory* $\mathcal C\_{ad}$ is the full fusion subcategory of $\mathcal C$ generated by $X\otimes X^\*$, where $X$ is simple.


$Rep(D(S\_3))$ has eight simple objects $\{1,\epsilon, \phi\_{i=1,\ldots,4},\psi\_\pm\}$ and $Rep(D(S\_3))\_{ad}$ is the subcategory generated by $\{1,\epsilon,\phi\_{i=1,\ldots,4}\}$. Its Grothendieck ring is commutative and determined by


$$
\begin{align\*}
\epsilon \otimes \epsilon &\cong 1 \\
\epsilon \otimes \phi\_i &\cong \phi\_i \\
\phi\_i \otimes \phi\_i &\cong 1 \oplus \epsilon \oplus \phi\_i \\
\phi\_i \otimes \phi\_j &\cong \phi\_k \oplus \phi\_l & i\neq j \neq k \neq l \\
\end{align\*}
$$


$Rep(D(S\_3))$ is modular so $Rep(D(S\_3))\_{ad}$ is braided(properly premodular). $Rep(D(S\_3))\_{ad}$ also admits a braiding with S-matrix


$$
\left(
\begin{array}{cccccc}
1 & 1 & 2 & 2 & 2 & 2 \\
1 & 1 & 2 & 2 & 2 & 2 \\
2 & 2 & 4 & 4 & 4 & 4 \\
2 & 2 & 4 & 4 & 4 & 4 \\
2 & 2 & 4 & 4 & 4 & 4 \\
2 & 2 & 4 & 4 & 4 & 4 \\
\end{array}\right).
$$


This braiding is symmetric $(s\_{ab}=d\_a d\_b)$ and so $Rep(D(S\_3))\_{ad}$ is equivalent as a fusion category to $Rep(G)$ for some finite group $G$. What is this $G$?



For what $G$ is $Rep(D(S_3))_{ad}$ Grothendieck equivalent to $Rep(G)$?
Despite my love of the finite group game, let me give an argument that doesn't use the classification of groups of order 18. The 1-dimensional objects correspond to representations of the abelianization. So your group must have abelianization $C\_2$, and so its commutator subgroup must be a group of size 9. Note that this splits as a semidirect product because there's a 2-sylow subgroup. 


A full tensor subcategory of $\mathrm{Rep}(G)$ which is closed under summands must be of the form $\mathrm{Rep}(G/N)$. (The proof of this is roughly the same as the proof that faithful representations tensor generate.) Thus your group must have $S\_3$ as a quotient in four different ways. Hence the commutator subgroup must be elementary abelian and the $C\_2$ must act on each of the factors by inversion (and thus on the whole thing by inversion).


