The following is a somewhat well-known fact: Given an even lattice $L$ with the pairing $\langle,\rangle: L\times L\to \mathbb{Z}$, we extend the pairing to $L\otimes \mathbb{Q}$ by tensoring with $\mathbb{Q}$. Then the discriminant group $A=L^*/L$ comes with a non-degenerate quadratic form $q: A\to \mathbb{Q}/\mathbb{Z}$ given by $q(x+L)=\langle x,x\rangle/2$. Conversely, any such pair $(A,q)$ is known to come from some even lattice $L$, see e.g. Wall, "Quadratic forms on finite groups".

My question is the $G$-symmetric case. Namely, suppose there is a finite abelian group $A$ with a non-degenerate quadratic form $q$ which is invariant under a subgroup $G$ on $Aut(A)$. Is there always a lattice $L$ with the pairing $\langle,\rangle$ which is invariant under $G\subset Aut(L)$ which gives rise to $(A,q)$?

(My motivation is to study the obstructions assocaited to $G$ actions on the modular tensor category associated to the metric group $(A,q)$, but I guess there can be various other applications, if the question I asked has a positive answer.)