Matrix logarithm for d-dimensional cyclic permutation matrix I want to find the matrix $\hat{H}_d$ which, when exponentiated, leads to a d-dimensional cyclic permutation transformation matrix.

I have solutions for d=2:
$$
    \hat{U}_2 =\left( \begin{matrix}
    0 & 1 \\
    1 & 0 \\
   \end{matrix} \right)=-i \exp\left(i\hat{H_2}\right) \to \\
    \hat{H}_2 =\left( \begin{matrix}
    0 & 1 \\
    1 & 0 \\
   \end{matrix} \right)
$$

d=3:
$$
    \hat{U}_3 =\left( \begin{matrix}
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    1 & 0 & 0\\
   \end{matrix} \right)=-i \exp\left(i\hat{H_3}\right) \to \\
    \hat{H}_3 =\frac{\pi}{3}\left( \begin{matrix}
    -\frac{1}{2} & \left(-\frac{1}{\sqrt{3}} + i \right) & \left(\frac{1}{\sqrt{3}} + i \right) \\
    \left(\frac{1}{\sqrt{3}} + i \right) & -\frac{1}{2} & \left(-\frac{1}{\sqrt{3}} + i \right)\\
    \left(\frac{1}{\sqrt{3}} + i \right) & \left(-\frac{1}{\sqrt{3}} + i \right) & -\frac{1}{2}
   \end{matrix}  \right)
$$

d=4:
$$
    \hat{U}_4 =\left( \begin{matrix}
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\
    1 & 0 & 0 & 0 \\
   \end{matrix} \right)=-i \exp\left(i\hat{H_4}\right) \to \\
    \hat{H}_4 =\frac{\pi}{4}\left( \begin{matrix}
    i & (1+i) & -i & (-1+i) \\
     (-1+i) & i & (1+i) & -i\\
     -i & (-1+i) & i & (1+i)\\
    (1+i) & -i & (-1+i) & i \\
   \end{matrix}  \right)
$$

Unfortunately, I am missing a general form of $H_d$ for $d>4$, so my question is:

Question: How does $\hat{H}_d$ look when I want that $\hat{U}_d=-i \exp\left(i \hat{H}_d \right)$ is a d-dimensional cyclic permutation transformation?


The article below presents the general form:
https://arxiv.org/abs/2001.11909
