Finding a suitable matrix to solve equation I'm trying to find a matrix $T$ such that it satisfies the following:
$$\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} =T^{-1} \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} T$$
As well as: $$\begin{bmatrix}e^{{2 \pi i}/3} & 0\\0 & e^{{4 \pi i}/3}\end{bmatrix} =T^{-1} \begin{bmatrix}\cos(2\pi/3) & -\sin(2\pi /3)\\ \sin(2\pi /3) & \cos(2\pi /3)\end{bmatrix} T$$
 A: $$T=\pmatrix{i&1\\1&i}$$ satisfies both equations.
A: Let $P=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$ and note that we have
$PT=TP$. If we let $T=\begin{bmatrix}a &b\\c&d\end{bmatrix}$, we see that
this gives $a=d,c=b$, so $T$ has the form $T=\begin{bmatrix}a &b\\b&a\end{bmatrix}$.
If $T$ is a solution it is easy to see that $\lambda T$ is a solution for
$\lambda \neq 0$, so we can take $T$ to have the form
$T = \begin{bmatrix}1 &t\\t&1\end{bmatrix}$.
The second equation can be written as
$T \begin{bmatrix}e^{{{2 \pi i} \over 3}} & 0\\0 & e^{{{4 \pi i} \over 3}}\end{bmatrix} =\begin{bmatrix}\cos({{2 \pi i} \over 3}) & -\sin({{2 \pi i} \over 3})\\ \sin({{2 \pi i} \over 3}) & \cos({{2 \pi i} \over 3})\end{bmatrix} T$. Computing the $(1,1)$ entry gives
$e^{{{2 \pi i} \over 3}} = \cos({{2 \pi i} \over 3}) -t \sin({{2 \pi i} \over 3})$, hence we must have $t=-i$.
Substituting and multiplying shows that $T$ satisfies the second equation (noting that $e^{{{4 \pi i} \over 3}} = e^{-{{2 \pi i} \over 3}}$).
As noted above, any non zero multiple of $T$ is a solution.
