Let us consider, for example, $\mathfrak{so}(6)$, with fifteen generators defined as (note: it sometimes has an $i$ in front)
$$\left(M_{ij}\right)_{ab}=-\left(\delta_{ia}\delta_{jb}-\delta_{ib}\delta_{ja}\right),\qquad i,j,a,b=1,\dots,6\,,\tag{1}$$
whose algebra is
$$\left[M_{ij},M_{k\ell}\right]=-\left(\delta_{i\ell}M_{jk}+\delta_{jk}M_{i\ell}-\delta_{ik}M_{j\ell}-\delta_{j\ell}M_{ik}\right).\tag{2}$$
It is usual (actually, I dont know if this is the unique option, that's why I said usual) to consider $M_{12}$, $M_{34}$ and $M_{56}$ ($H_i=M_{(2i-1)(2i)}$, in general) as the Cartan generators. The positive simple roots in this case are found to be (page 84):
$$ E_{\alpha_1}=M_{35}-i M_{36}+i M_{45}+M_{46},\\ E_{\alpha_2}=M_{13}-i M_{14}+i M_{23}+M_{24},\\ E_{\alpha_3}=M_{13}-i M_{14}-i M_{23}-M_{24}, \tag{3}$$
My question is if it is possible to have another set of Cartan generators, e.g. $M_{16}$, $M_{25}$ and $M_{34}$ or a combination of them. How could I get this second set of "Cartans" from the first one? Does the transformation also apply for the roots?
My second question is about finding the roots in terms of the generators $M_{ij}$ for the case of $\mathfrak{so}(6)$ whose generators are written in such a way they show the nine generators of $\mathfrak{u}(3)$. The $\mathfrak{u}(3)$ subalgebra of $\mathfrak{so}(6)$ is composed by those generators of $\mathfrak{so}(6)$ that are invariant under $J=\begin{pmatrix} 0 & I_3 \\ -I_3 & 0\end{pmatrix}$. Thus, those of $\mathfrak{so}(6)/\mathfrak{u}(3)$ are
$$ K_1=\frac{1}{2}\left(M_{12}-M_{45}\right),\quad K_2=\frac{1}{2}\left(M_{15}-M_{24}\right),\quad K_3=\frac{1}{2}\left(M_{13}-M_{46}\right),\\ K_4=\frac{1}{2}\left(M_{16}-M_{34}\right),\quad K_5=\frac{1}{2}\left(M_{26}-M_{35}\right),\quad K_6=\frac{1}{2}\left(M_{23}-M_{56}\right), \tag{4}$$
and those of $\mathfrak{u}(3)$,
$$ T_1=\frac{1}{2}\left(M_{12}+M_{45}\right),\quad T_2=\frac{1}{2}\left(M_{15}+M_{24}\right),\quad T_3=\frac{1}{2}\left(M_{13}+M_{46}\right),\\ T_4=\frac{1}{2}\left(M_{16}+M_{34}\right),\quad T_5=\frac{1}{2}\left(M_{26}+M_{35}\right),\quad T_6=\frac{1}{2}\left(M_{23}+M_{56}\right),\\ T_7=M_{14},\quad T_8=M_{25},\quad T_9=M_{36}. \tag{5}$$
Notice that the first Cartans, $M_{12}$, $M_{34}$ and $M_{56}$, and the second ones, $M_{16}$, $M_{25}$ and $M_{34}$, appear mixed in this representation. In particular, for my purposes, I need the latter set as being the Cartans or included in them.
Summarizing: what is the operation to "rotate" the Cartan generators and roots to another set of them? And, in general, given a set of generators, how do I get the new Cartan generators and roots?
I know that for $\mathfrak{so}(6)$ the Cartan matrix and subalgebra are
$$ A_{ij}=\begin{pmatrix} 2 & -1 & -1\\ -1 & 2 & 0 \\ -1 & 0 & 2 \end{pmatrix},\quad \left[H_i,E_j\right]=A_{ij}E_j,\quad \left[E_i,E_{-j}\right]=\delta_{ij}H_j \tag{6}$$
but I dont know how to implement an efficient way to "extract" the roots, now, in terms of $K_i$ and $T_i$.