Let's say that $M,N$ are two invertible $n\times n$ integer matrices, such that all their eigenvalues are greater than $1$ in magnitudes. Then we define $$K_M:=(M^{-1}\mathbb{Z}^d)\cap[0,1)^n,\quad K_N=(N^{-1}\mathbb{Z}^d)\cap[0,1)^n.$$ Then $K_M$ is precisely the complete set of representatives of distinct cosets of the quotient group $(M^{-1}\mathbb{Z}^n)/\mathbb{Z}^n$, and similarly for $K_N$.
Now let's consider $$K_{M,N}:=K_M\cap K_N.$$ If $n=1$, then we see that $M,N$ are integers with magnitudes greater than $1$. Then it's not difficult to see that $d\in K_{M,N}$ if and only if $d$ is of the form
$$\frac{j}{gcd(M,N)},\quad j=0,\dots,gcd(M,N)-1.$$ Thus in this case $$K_{M,N}=K_{gcd(M,N)}=(gcd(M,N)^{-1}\mathbb{Z})\cap[0,1).$$ My question is, what happens if the dimention $n\geq 2$? can we always find an invertible integer matrix $A$ such that $K_M\cap K_N=K_A$? Here $K_A=(A^{-1}\mathbb{Z}^n)\cap[0,1)^n$.
What I thought is the following: let $$G_M:=(M^{-1}\mathbb{Z}^n)/\mathbb{Z}^n,\quad G_N:=(N^{-1}\mathbb{Z}^n)/\mathbb{Z}^n.$$ Denote
$$\Omega_{M,N}:=\{k+\mathbb{Z}^d:k\in K_{M,N}\}.$$ Then we can view $\Omega_{M,N}$ as a normal subgroup of $G_M$, and also a normal subgroup of $G_N$. Now the question becomes: is there an invertible integer matrix $A$ such that $(A^{-1}\mathbb{Z}^n)/\mathbb{Z}^n=\Omega_{M,N}$?
At this point I cannot proceed further. Any help will be greately appreciated.