A conjugation problem Let  $A$ and $B$ be $2\times 2$ matrices with $\mathrm{tr}(A)>2$, $\mathrm{tr}(B)>2$ and $\det(A)=1$, $\det(B)=1$.
My Question : there exists an bijetive application $F:\mathbb{R}^{2}\to\mathbb{R}^{2}$ such that  $F(\mathbb{Z}^2)=\mathbb{Z}^2$ and $A\circ F=F\circ B$??
EDIT 1:
Think  $A$, $B$ and $C$ as applications of $\mathbb{R}^2\to \mathbb{R}^2$
EDIT2: $A$ and $B$ has integer entries.
I apologize for the careless drafting the question
 A: If your $F$ exists and is linear, then $A=F^{-1}BF$, so $A$ and $B$ would have the same eigenvalues. This means that any $A$, $B$ with different eigenvalues give a counterexample. For instance, take
$$
A=\begin{bmatrix}3&1\\ 2 &1\end{bmatrix}, \ \ \ B=\begin{bmatrix}4&1\\3&1\end{bmatrix}.
$$
Then for any $F$ like you want, $AF$ has eigenvalues $2\pm\sqrt3$, while $FB$ has eigenvalues $(5\pm\sqrt{21})/2$, so they cannot be equal.
Note that in this example $F$ does not exist even without the requirement of integer entries. 
A: 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 $(*).$
