A matrix $A\in\textbf{Mat}\_n(\mathbb{R})$ is called asymptotically nilpotent if for each vector $v$, ${\lim}\_{k\to\infty}A^k(v) = 0$. Assume that $\mathcal{A}, \mathcal{B}$ be maximal (under inclusion) among those subsets of $\textbf{Mat}\_n(\mathbb{R})$ with only asymptotically nilpotent matrices and which are closed under the Lie bracket operation (i.e. $\forall X, Y\in\mathcal{A}, [X, Y] = XY - YX\in \mathcal{A}$ and similarly for $\mathcal{B}$).



> 
> Is it true that for some $P\in\text{GL}\_n(\mathbb{R})$,
> $P\mathcal{A}P^{-1} = \mathcal{B}$?
> 
> 
> 


Further variation of this question is asked [here](https://mathoverflow.net/questions/432547/asymptotically-nilpotent-lie-sets-of-matrices).



Asymptotically nilpotent matrices
The answer to your first question is no: they're not all conjugate.


Indeed, let $A$ be the set of all upper triangular matrices of absolute value $<1$ on the diagonal. Then $A$ consists of asymptotically nilpotent matrices (clear) and is maximal for this property (1).


Let $B\_0$ be the set of all $d\times d$ matrices with all coefficients of absolute value $<1/2d^2$. Then $B\_0$ is stable under taking brackets and consists of asympotically nilpotent matrices. Let $B$ be maximal for these properties and containing $B\_0$. Then $B$ is not conjugate to $A$, since $0$ is in the interior of $B$ and not of $A$.


(1) Let $A'$ be a larger subset with the given properties, and let by contradiction $M\in A'$ be a matrix outside $A$. If $M$ is upper triangular, then $M$ has a diagonal coefficient of absolute value $\ge 1$ and hence is not asymptotically nilpotent. Otherwise $M\_{ji}\neq 0$ for some $i<j$. Choose $(i,j)$ with $j-i$ maximal for this property, and then among the possible choices, choose $i$ minimal. Write $M'=[M,E\_{ij}]$. Then by stability, $[M,tE\_{ij}]=tM'$ belongs to $A'$, for every scalar $t$. By the choices, $M'$ is upper triangular, and $M'\_{ii}\neq 0$. This is a contradiction.


