Applying the modularisation/deequivariantisation procedure to the representation category $\operatorname{Rep}_G$ of a finite group $G$ with trivial braiding gives the fibre functor to vector spaces. What happens if we chose a nontrivial braiding?

My vague understanding from this question is that we need to choose an abelian normal subgroup $A \hookrightarrow G$ together with a bilinear form on the function algebra $\beta: \mathbb{C}(A) \otimes \mathbb{C}(A) \to \mathbb{C}^*$ to get a quasitriangular structure on the group algebra $\mathbb{C}[G]$. Supposedly, a skew-linear $\beta$ leads to a triangular braiding (which I don't find obvious, and I can't find a detailed proof).

Can the modularisation functor somehow be expressed as composition $\iota^*$ with a subgroup inclusion $\iota: A' \hookrightarrow G$, and $A'$ another subgroup to be determined from $A$ and $\beta$? If not, how does the modularisation relate to the given data (on which it must depend completely)?

Edit: From Qiaochu Yuan's excellent answer to my question, I can add a little more to my vague conjecture.

The symmetric centre of $\operatorname{Rep}_G$ is a symmetric fusion category. Let's assume for a moment that its twist is trivial (i.e. no supergroups), then we can restrict the fibre functor of $\operatorname{Rep}_G$ to the symmetric centre,so it's the representations of another group, which we'll call $G'$. Intuitively, it should be the group that we are going to deequivariantise away!

We can restrict the automorphisms of the fibre functor of $\operatorname{Rep}_G$ to the fibre functor of $\operatorname{Rep}_{G'}$, which gives us a surjective group homomorphism $\phi: G \twoheadrightarrow G'$ such that $\phi^*: \operatorname{Rep}_{G'} \hookrightarrow \operatorname{Rep}_G$ is the inclusion of the symmetric centre. My conjecture is now that $A'$ is the kernel of $\phi$!