The following \(2 \times 2\) matrix is very interesting:
\begin{equation*}
J = \frac{1}{\sqrt{2}} \begin{bmatrix}
1 & i \\
1 & -i
\end{bmatrix}
\end{equation*}
It satisfies the following identities:
\begin{align*}
J^{3} &= \frac{1}{\sqrt{2}}(1 + i) I & J^{6} &= i I & J^{9} &= -\frac{1}{\sqrt{2}}(1 - i) I & J^{12} &= -I \\
J^{15} &= -\frac{1}{\sqrt{2}}(1 + i) I & J^{18} &= -i I & J^{21} &= \frac{1}{\sqrt{2}}(1 - i) I & J^{24} &= +I
\end{align*}
Note that \(J\) is unitary:
\begin{equation*}
J^{\dagger} = \frac{1}{\sqrt{2}} \begin{bmatrix}
1 & 1 \\
-i & i
\end{bmatrix} = J^{23} = J^{-1}
\end{equation*}
Thus
\begin{align*}
J^{4} &= \frac{1}{\sqrt{2}}(1 + i) J & J^{7} &= i J & J^{10} &= -\frac{1}{\sqrt{2}}(1 - i) J & J^{13} &= -J \\
J^{16} &= -\frac{1}{\sqrt{2}}(1 + i) J & J^{19} &= -i J & J^{22} &= \frac{1}{\sqrt{2}}(1 - i) J
\end{align*}
and
\begin{align*}
J^{2} &= \frac{1}{\sqrt{2}}(1 + i) J^{\dagger} & J^{5} &= i J^{\dagger} & J^{8} &= -\frac{1}{\sqrt{2}}(1 - i) J^{\dagger} & J^{11} &= -J^{\dagger} \\
J^{14} &= -\frac{1}{\sqrt{2}}(1 + i) J^{\dagger} & J^{17} &= -i J^{\dagger} & J^{20} &= \frac{1}{\sqrt{2}}(1 - i) J^{\dagger}
\end{align*}
Note that the coefficients correspond to the 8th-roots of unity, reflecting the fact that the cyclic group \(\mathbb{Z}_{24}\) is the product group \(\mathbb{Z}_{3} \times \mathbb{Z}_{8}\).
If the kets \(|0 \rangle\) and \(|1 \rangle\) are defined via
\begin{align*}
|0 \rangle &= \begin{bmatrix}
1 \\ 0
\end{bmatrix} & |1 \rangle &= \begin{bmatrix}
0 \\ 1
\end{bmatrix}
\end{align*}
and the kets \(|Y{-}\rangle\) and \(|Y{+}\rangle\) are defined via
\begin{align*}
|Y{-}\rangle &= \frac{1}{\sqrt{2}} \left( |0 \rangle - i | 1 \rangle \right) & |Y{+}\rangle &= \frac{1}{\sqrt{2}} \left( |0 \rangle + i | 1 \rangle \right)
\end{align*}
then \(J\) and \(J^{\dagger}\) transform between these two ortho-normal bases:
\begin{align*}
J^{\dagger} |0 \rangle &= |Y{-} \rangle & J^{\dagger} |1 \rangle &= |Y{+} \rangle
\end{align*}
Note that the kets \(|Y{-}\rangle\) and \(|Y{+}\rangle\) are eigen-vectors of the unitary hermitian matrix \(Y\) given by
\begin{align*}
Y |Y{-}\rangle &= -|Y{-}\rangle & Y |Y{+}\rangle &= |Y{+}\rangle & Y &= \begin{bmatrix}
0 & -i \\
i & 0
\end{bmatrix}
\end{align*}
This is one of the Pauli matrices.
Let \(J_{2} = J\). It follows that
\begin{equation*}
(J_{2})^{24} = I_{2}
\end{equation*}
Note that \(J_{2}\) correspond to a 1-qubit gate. Consider the Kronecker product:
\begin{equation*}
J_{4} = J_{2} \otimes J_{2}
\end{equation*}
It follows that
\begin{equation*}
(J_{4})^{12} = I_{4}
\end{equation*}
Note that \(J_{4}\) correspond to a 2-qubit gate. Now consider the Kronecker product:
\begin{equation*}
J_{16} = J_{4} \otimes J_{4}
\end{equation*}
It follows that
\begin{equation*}
(J_{16})^{6} = I_{16}
\end{equation*}
Note that \(J_{16}\) correspond to a 4-qubit gate. Now consider the Kronecker product:
\begin{equation*}
J_{256} = J_{16} \otimes J_{16}
\end{equation*}
It follows that
\begin{equation*}
(J_{256})^{3} = I_{256}
\end{equation*}
Note that \(J_{256}\) correspond to an 8-qubit gate.
