Given diagonal Matrix and number $q$, find nilpotent matrix such that $D N D^{-1} = qN$. Given a diagonal matrix $D$ and an integer $q$, find a nilpotent matrix $N$ such that $$DND^{-1}= qN$$ holds. 
Just thinking about this for a bit, I'm not convinced it's possible, since the multiplication by $D \cdot D^{-1}$ gives a new matrix with entries $$b_{ij} = \frac{d_{ii}}{d_{jj}} n_{ij},$$ or maybe it's $$b_{ij} = \frac{d_{jj}}{d_{ii}} n_{ij}$$...
Either way, the resulting matrix $B$ has the same diagonal as $N$ and differs elsewhere by different constant factors. So, then is it possible to find such a nilpotent matrix? I tried looking at upper and lower triangular matrices but to no avail.
A hint would be welcome if it indeed can be solved.
 A: Indeed in $2-$D, the only nontrivial nilpotent matrix is $\left(\begin{array}{cc}0 & \epsilon\\ 0 & 0 \end{array}\right)$, $\epsilon\neq0$, (or it's transpose), and the equality can hold iff $q=\frac{d_{ii}}{d_{jj}}$. So this argument suggests it cannot be done.
Next, why not suppose that $N$ is upper triangular. Since $N$ has all zeros on the diagonal, then $N^{d-1}$ has $n_{1d}$ the only possible nonzero entry, and then since $DN^{d-1}D^{-1}=(DND^{-1})^{d-1}=q^{d-1}N^{d-1}$, again it must be that this holds iff $q^{d-1}=\frac{d_{ii}}{d_{jj}}$.
The only part I'm not 100% on is dealing with the general case: we can always triangularize any matrix $N$ (this is Schur's triangularization theorem), but of course the diagonal matrix $D$ will probably not be diagonal in the basis that triangularizes $N$...
EDIT: For the general case: write $N=TUT^{-1}$, where $U$ is upper triangular (again, with zeros on the diagonal) with $U^{d-1}$ having only the top-right entry nonzero. Replace this in the equality above
$$$DTU^{d-1}(DT)^{-1}=(DTN(DT)^{-1})^{d-1}=q^{d-1}TU^{d-1}T^{-1}\Longrightarrow q^{d-1}U^{d-1}=QU^{d-1}Q^{-1}$$
for $Q=T^{-1}DT$. Thus given $D$ this equality will restrict what $q$ can be. You might be able to explicitly write down what $q$ must be given $D$ and some change of basis $T$, but since $U^{d-1}$ is a single entry in the $1d$ spot, this shows that this won't work for any general $D,q$. 
