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?