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"](https://mathscinet.ams.org/mathscinet-getitem?mr=322071).


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.)



Do $G$-invariant non-degenerate quadratic forms come from $G$-invariant even lattices?
The group of automorphisms of $A$ which preserve the quadratic form $q$ is known as the orthogonal group $O(A,q)$. Likewise, if $L$ is a free $\mathbb{Z}$-module of finite rank with an even $\mathbb{Z}$-valued quadratic form $Q$, there is an orthogonal group $O(L,Q)$. When $(A,q)$ is the discriminant form of $(L,Q)$, there is a natural homomorphism $O(L,Q) \to O(A,q)$.


It is a consequence of the Strong Approximation Theorem for the orthogonal group that this homomorphism is surjective when $L$ is indefinite and of sufficiently large rank. (I believe that this is proven in several of the standard books about integral quadratic forms.) Note that adding a copy of the hyperbolic plane (resp. the $E\_8$ lattice) to $L$ does not change the disriminant form, but makes $L$ indefinite (resp. increases its rank). Thus, if we are given $(A,q)$, an $L$ can be found with this property.


This doesn't quite answer your question, because you want to know if the surjection $O(L,Q) \to O(A,q)$ can be split. I don't know, but I am inclined to doubt it. Consider the analogous problem for the special linear group instead of the orthogonal group. We have surjections $SL(n,\mathbb{Z}) \to SL(n,\mathbb{F}\_p)$ which cannot be split for $p$ sufficiently large.


