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Given two matrices $A, B \in \mathcal{M}_n(\mathbb{Z})$ with same characteristic polynomial $f(t)$, it is difficult to know when they are conjugate (i.e. there exists $U\in GL_n(\mathbb{Z})$ such that $B=UAU^{-1}$). We know also that such $A$ and $B$ are conjugate over $\mathbb{Q}$ (i.e. there exists $V\in GL_n(\mathbb{Q})$ such that $B=VAV^{-1}$).

Do you know some kind of matrices $W\in GL_4(\mathbb{Q})\setminus GL_4(\mathbb{Z})$ which ensures that $A$ and $WAW^{-1}$ are not conjugates over $\mathbb{Z}$?

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    Since $WIW^{-1}=I=I(I)I^{-1}$, no such $W$ exists. – 2017-01-09
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    Sorry, I suppose that $f(t)\in \mathbb{Z}[t]$ irreducible (even over $\mathbb{Q}$) and with degree $4$, so $A\neq I$. – 2017-01-09

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I think there is no general form of $W$ such that $A$ and $WAW^{-1}$ are not conjugated in $GL_n(\mathbb{Z})$ for arbitrary $A\in M_n(\mathbb{Z})$. What we can say is, that for some $A$ the only possibilities for $W$ have determinant $\neq \pm 1$, because of its Jordan form. Hence $A$ and $WAW^{-1}$ cannot be conjugated in $GL_n(\mathbb{Z})$. Consider a simple example, namely two matrices $A,B\in GL_2(\mathbb{Z})$, $$ A=\begin{pmatrix} 5 & 2 \cr 2 & 1\end{pmatrix},\;B=\begin{pmatrix} 5 & 4 \cr 1 & 1\end{pmatrix}, $$ which have both the characteristic polynomial $\chi(t)=t^2-6t+1$, but do not admit a $W\in GL_2(\mathbb{Z})$ with $B=WAW^{-1}$, because the Jordan form of any $W$ with $AW=WB$ is given by $$ J_W=\begin{pmatrix} a & 2b \cr b & 2(a-2b)\end{pmatrix}, $$ which has determinant a multiple of $2$, and hence is not $1$ or $-1$ as requested (but for the question of similarity or non-similarity I consider the matrix $W$ with integral coefficients). From these $2\times 2$ examples, we can construct $4\times 4$-examples by using blocks.