Jordan matrix (de)composition So any square matrix $A$ can be decomposed into $A = S J S^{-1}$ where $J$ has a normal Jordan form, moreover $A$ and $J$ are similar matrices.
My question is quite straightforward. Given arbitrary normal Jordan matrix $J$, can I always find $S$ such that the resulting matrix $A$ has real integer entries?
Edit: So we have some necessary conditions now namely that the diagonal elements of $J$ must be real or come in complex conjugate pairs.
It is still unclear though what are other conditions and most importantly what is a reasonable approach in order to actually calculate some $A$ given $J$.

Besides general approach I'm also considering a particular (not overly complicated I believe) case
$$
S 
\begin{pmatrix}
i & 1 & 0 \\ 0 & i & 0 \\ 0 & 0 & i
\end{pmatrix}
S^{-1} = A \in \mathbb{Z}^{3 \times 3},
$$
but I'm not sure how to proceed. (Proved impossible, see edit.)
 A: The answer depends. The diagonal elements of $J$ are eigenvalues of $A$. So if $A$ is real, its characteristic polynomial also has real coefficients, and therefore the eigenvalues come in complex conjugate pairs. the $3\times 3$ case you give obviously does not satisfy.
The whole problem can be seen as deciding whether the orbit of $J$ under similarity transformation of $S\in GL(n,\mathbb C)$ has intersection with $\mathbb Z^{n\times n}$. This is simply beyond my knowledge. Wiki seems to show little information.
A: It seems the following.
The obvious necessary condition is 
(C) the characteristic polynomial $\det (\lambda I-J)$ of the matrix $J$ has integer coefficients. 
If all roots of the polynomial $\det (\lambda I-J)$ are mutually different, then Condition C is also sufficient. Indeed, in this case the matrix $J$ is diagonal, and each matrix $A$ with $\det (\lambda I-A)=\det (\lambda I-J)$ has Jordan normal form $J$.  So it suffices to find a matrix $A$ with integer entries such that $\det (\lambda I-A)=\det (\lambda I-J)$, which is easy to do. Indeed, if  $$\det (\lambda I-J)=a_0+a_1\lambda+\dots+a_{n-1}\lambda^{n-1}+\lambda^n$$ then we can put
$$A=\begin{pmatrix} 
0 & 0 & \dots & 0 & -a_0 \\ 
1 & 0 & \dots & 0 & -a_2 \\ 
0 & 1 & \dots & 0 & -a_3 \\ 
\dots & \dots & \dots & \dots & \dots \\
0 & 0 & \dots & 0  & -a_{n-2} \\ 
0 & 0 & \dots & 1  & -a_{n-1} 
\end{pmatrix}.$$
Another sufficient condition is when all roots of the polynomial $\det (\lambda I-J)$ are already integers.
Consider the case when order $n$ of the matrix $J$ equals $2$. If its characteristic polynomial $\det (\lambda I-J)$ has different roots, then Condition C is both necessary and sufficient. Assume that the polynomial $\det (\lambda I-J)$ has only one root $\lambda_0$. Then $\det (\lambda I-J)=(\lambda-\lambda_0)^2$. If Condition C is satisfied, then both $2\lambda_0$ and $\lambda_0^2$ are integers. So $\lambda_0$ is a integer too, and thus the matrix $J$ has integer entries. So in this case Condition C is both necessary and sufficient too. 
Similarly, if the matrix $J$ has a unique root $\lambda_0$ and Condition C is satisfied, then both $n\lambda_0$ and $\lambda_0^n$ are integers, so $\lambda_0$ is an integer too, and thus the matrix $J$ has integer entries. So in this case Condition C is both necessary and sufficient too. 
The further steps in this investigation are to consider case $n=3$ and to find a counterexample of the matrix $J$  satisfying Condition $C$, but which is not similar to a matrix $A$ with integer entries. 
PS. To be continued...
