A parcial solution : I could only show that
$F|_{\mathbb{Z}^2}\circ A|_{\mathbb{Z}^2}=B|_{\mathbb{Z}^2}\circ F|_{\mathbb{Z}^2} ~~~~~~(*)$
but this is enough for my purposes.
In fact, consider in $\mathbb{Z}^2$ the equivalence relation $\sim_A$ $\underline{x} \sim_A \underline{y}~~~~ \text{if only if} ~~~~Orb_A(\underline{x})=Orb_A(\underline{y})$
and similarly define the equivalence relation $\sim_B.$ Let $G_A:=\mathbb{Z}^2 /\sim_A$ and $G_B:=\mathbb{Z}^2 /\sim_B$.
$G_A$ and $G_B$ are infinite enumerable sets, then there is a bijection $\Lambda$ between them.
Now we show how to construct $ F $ over each orbit of $ A $. Let be, $Orb_A(\underline{x})$ and $Orb_B(\underline{y})$ such that
$\Lambda(Orb_A(\underline{x}))=Orb_B(\underline{y})~~~~$ define $~~F(\underline{x})=\underline{y}, ~~F(A\underline{x})=B\underline{y}, \ldots F(A^k\underline{x})=B^k\underline{y}, \text{etc.}$
See, $ F $ is bijetive along each orbit, since neither nor $ A $ nor $ B $ have eigenvalues which are the roots of unity
It is easy to see that $ F $ so defined satisfies $(*).$