In this question I want to consider a quadratic form $\mathcal{Q}$ on an even integral finite rank lattice (not necessarily self-dual, unimodular, or positive definite).
Given an explicit such quadratic form, are there elementary methods to write down exactly what the full symmetry group is? If so, say one has found a subgroup of the full symmetry group, is there a way to be sure that in fact the complete symmetry group has been found?
One example of interest to me is the following. Consider the lattice $\Gamma =\mathbb{Z}^{3}$ endowed with quadratic form,
$$\mathcal{Q}(a,b,c) = 2(ab + ac + bc)$$
I found a group representation of $\text{PGL}_{2}(\mathbb{Z})$ on the lattice automorphisms of $\Gamma$ defined by the following three automorphisms:
$$\sigma_{A}(a,b,c) = (b,a,c), \,\,\,$$ $$\sigma_{B}(a,b,c) = (a+2c, b+2c, -c), \,\,\,$$ $$\sigma_{C}(a,b,c) = (a+2b, 2b+c, -b).$$
(The above are the images of the three usual generators of $\text{GL}_{2}(\mathbb{Z})$.) One can show easily that the three automorphisms above preserve the quadratic form $\mathcal{Q}$. Therefore, whatever the full symmetry group of $\mathcal{Q}$ is, I believe it at least contains $\text{PGL}_{2}(\mathbb{Z})$ as a subgroup. Now in this case I'm nearly certain the full symmetry group is much larger, so as indicated in my question above, how can I determine what the rest of it is, if indeed this is possible for this form?
