Null-spaces modulo n Let A and B be square integer-valued matrices (possibly of different size) such that their null-spaces modulo n (i.e. the set of vectors v that satisfy Av = 0 mod n) are isomorphic (i.e. there exists a bijection f with f(0)=0 and f(x+y)=f(x)+f(y)) for all n. Moreover, there exist functions F and G that work as isomorphisms for all n (with F∘G=id on the null-spaces). Can we show that the determinants of A&B are equal up to a sign?
For context, this is from knot theory. I am trying to prove that the determinants of the Goeritz and the colouring matrix are equal up to a sign, and so far I've got this.
 A: 
Claim: The null space of the size $k$ square matrix $A$ modulo $n$ is isomoprhic (as an abelian group) to
$$
(\Bbb Z / \gcd (\alpha_1,n) \Bbb Z) \times  \cdots \times (\Bbb Z/ \gcd (\alpha_k,n) \Bbb Z),
$$
where $\alpha_1,\dots,\alpha_k$ are the invariant factors of $A$.

Proof: Let $D$ denote the Smith normal form of $A$, and let $U,V$ be unimodular  matrices (integer matrices with integer matrix inverses) such that $A = UDV$. Note that the null space of $A$ is equal to the nullspace of $U^{-1}A = DV$. Furthermore, the map $\phi(x) = Vx$ defines an isomorphism over $(\Bbb Z/n \Bbb Z)^k$ that restricts to a bijection from the nullspace of $DV$ to the nullspace of $D$.  Thus, it suffices to determine the nullspace of $D$ up to isomorphism.
For $x = (x_1,\dots,x_k) \in (\Bbb Z/n\Bbb Z)^k$, we compute
$$
Dx = (\alpha_1 x_1,\dots,\alpha_kx_k).
$$
We see that $x \in \operatorname{null}(D) \iff n \mid \alpha_i x_i$ for $i = 1,\dots, k$. Equivalently, the null space consists of the vectors $x$ for which $(n/\gcd(\alpha_i,n)) \mid x_i$ for each $i = 1,\dots,k$.
From there, note that the multiples of $n/\gcd(\alpha_i,n)$ modulo $n$ form a subgroup of $\Bbb Z/n \Bbb Z$ isomorphic to $\Bbb Z/\gcd(\alpha_i,n) \Bbb Z$.
