The standard 1-qubit information ortho-normal basis is
\begin{align*}
|b_{0} \rangle &= | 0 \rangle & |b_{1} \rangle &= | 1 \rangle
\end{align*}
The standard Bell ortho-normal basis is
\begin{align*}
|B_{20} \rangle &= \frac{| 0 \rangle + | 1 \rangle}{\sqrt{2}} & |B_{21} \rangle &= \frac{| 0 \rangle - | 1 \rangle}{\sqrt{2}}
\end{align*}
There are two ways to map the information basis to the Bell basis. The usual way involves the Hadamard transform:
\begin{equation*}
H_{2} = |B_{20} \rangle \langle 0 | + |B_{21} \rangle \langle 1 | = \frac{1}{\sqrt{2}} \begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}
\end{equation*}
That is
\begin{align*}
H_{2} |0 \rangle &= |B_{20} \rangle & H_{2} |1 \rangle &= |B_{21} \rangle
\end{align*}
Note that
\begin{equation*}
(H_{2})^{2} = I_{2}
\end{equation*}
The other way is
\begin{equation*}
J_{2} = |B_{21} \rangle \langle 0 | + |B_{20} \rangle \langle 1 | = \frac{1}{\sqrt{2}} \begin{bmatrix}
1 & 1 \\
-1 & 1
\end{bmatrix}
\end{equation*}
That is
\begin{align*}
J_{2} |0 \rangle &= |B_{21} \rangle & J_{2} |1 \rangle &= |B_{20} \rangle
\end{align*}
Note that
\begin{align*}
(J_{2})^{2} &= \begin{bmatrix}
0 & 1 \\
-1 & 0
\end{bmatrix} & (J_{2})^{4} &= -I_{2} & (J_{2})^{6} &= \begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix} & (J_{2})^{8} &= I_{2}
\end{align*}
Let
\begin{align*}
(J_{2})^{0} &= I_{2} & (J_{2})^{1} &= A & (J_{2})^{2} &= B & (J_{2})^{3} &= -A^{\dagger} \\
(J_{2})^{4} &= -I_{2} & (J_{2})^{5} &= -A & (J_{2})^{6} &= -B & (J_{2})^{7} &= A^{\dagger}
\end{align*}
That is, the powers of \(J_{2}\) form a representation of \(\mathbb{Z}_{2} \times \mathbb{Z}_{4}\).
Complex 1-Bell Basis
The ortho-normal complex Bell basis is
\begin{align*}
|C_{20} \rangle &= \frac{| 0 \rangle + i| 1 \rangle}{\sqrt{2}} & |C_{21} \rangle &= \frac{| 0 \rangle - i| 1 \rangle}{\sqrt{2}}
\end{align*}
Again, you have two mappings from the information basis to the complex Bell basis:
\begin{align*}
A_{2} &= |C_{20} \rangle \langle 0 | + |C_{21} \rangle \langle 1 | = \frac{1}{\sqrt{2}} \begin{bmatrix}
1 & 1 \\
i & -i
\end{bmatrix} & B_{2} &= |C_{21} \rangle \langle 0 | + |C_{20} \rangle \langle 1 | = \frac{1}{\sqrt{2}} \begin{bmatrix}
1 & 1 \\
-i & i
\end{bmatrix}
\end{align*}
Note that
\begin{equation*}
(A_{2})^{3} = I_{2} \exp\left( \frac{\pi i}{4} \right)
\end{equation*}
and also
\begin{equation*}
(B_{2})^{3} = I_{2} \exp\left( -\frac{\pi i}{4} \right)
\end{equation*}
The powers of each of these matrices form a representation of \(\mathbb{Z}_{3} \times \mathbb{Z}_{8}\).
Real 2-Bell Basis
The standard 2-qubit information ortho-normal basis is
\begin{align*}
|b_{0} \rangle &= |00 \rangle & |b_{1} \rangle &= |01 \rangle \\
|b_{2} \rangle &= |10 \rangle & |b_{3} \rangle &= |11 \rangle
\end{align*}
The standard 2-Bell ortho-normal basis is
\begin{align*}
|B_{40} \rangle &= \frac{1}{\sqrt{2}} \left( |00 \rangle + |11 \rangle \right) & |B_{41} \rangle &= \frac{1}{\sqrt{2}} \left( |00 \rangle - |11 \rangle \right) \\
|B_{42} \rangle &= \frac{1}{\sqrt{2}} \left( |01 \rangle + |10 \rangle \right) & |B_{43} \rangle &= \frac{1}{\sqrt{2}} \left( |01 \rangle - |10 \rangle \right)
\end{align*}
There are 24 possible mappings from the information basis to the Bell basis:
\begin{equation*}
H_{4}(w x y z) = |B_{4w} \rangle \langle 00 | + |B_{4x} \rangle \langle 01 | + |B_{4y} \rangle \langle 10 | + |B_{4z} \rangle \langle 11 |
\end{equation*}
Here \(w x y z\) is one of the permutations of the sequence 0123. These 24 possibilities fall into 5 categories. Two satisfy:
\begin{equation*}
\left[ H_{4}(0 2 3 1) \right]^{2} = \left[ H_{4}(2 0 1 3) \right]^{2} = I_{4}
\end{equation*}
Four satisfy:
\begin{equation*}
\left[ H_{4}(0 1 3 2) \right]^{4} = \left[ H_{4}(2 1 0 3) \right]^{4} = \left[ H_{4}(3 0 1 2) \right]^{4} = \left[ H_{4}(3 2 0 1) \right]^{4} = I_{4}
\end{equation*}
Four satisfy:
\begin{equation*}
\left[ H_{4}(1 0 3 2) \right]^{12} = \left[ H_{4}(2 1 3 0) \right]^{12} = \left[ H_{4}(2 3 0 1) \right]^{12} = \left[ H_{4}(3 0 2 1) \right]^{12} = I_{4}
\end{equation*}
Four satisfy:
\begin{equation*}
\left[ H_{4}(0 1 2 3) \right]^{24} = \left[ H_{4}(0 3 1 2) \right]^{24} = \left[ H_{4}(1 2 0 3) \right]^{24} = \left[ H_{4}(3 2 1 0) \right]^{24} = I_{4}
\end{equation*}
The rest satisfy
\begin{equation*}
\left[ H_{4}(w x y z) \right]^{8} = I_{4}
\end{equation*}
Let \(A_{2} \equiv H_{4}(0 2 3 1)\). Then the powers of \(A_{2}\) form a representation of \(\mathbb{Z}_{2}\). Let \(A_{4} \equiv H_{4}(0 1 3 2)\). Then the powers of \(A_{4}\) form a representation of \(\mathbb{Z}_{4}\). Let \(A_{8} \equiv H_{4}(0 2 1 3)\). Then the powers of \(A_{8}\) form a representation of \(\mathbb{Z}_{8}\). Let \(A_{12} \equiv H_{4}(1 0 3 2)\). Then the powers of \(A_{12}\) form a representation of \(\mathbb{Z}_{2} \times \mathbb{Z}_{6}\). Let \(A_{24} \equiv H_{4}(0 1 2 3)\). Then the powers of \(A_{24}\) form a representation of \(\mathbb{Z}_{2} \times \mathbb{Z}_{12}\).
Complex 2-Bell Basis
The complex 2-Bell ortho-normal basis is
\begin{align*}
|C_{40} \rangle &= \frac{1}{\sqrt{2}} \left( |00 \rangle + i|11 \rangle \right) & |C_{41} \rangle &= \frac{1}{\sqrt{2}} \left( |00 \rangle - i|11 \rangle \right) \\
|C_{42} \rangle &= \frac{1}{\sqrt{2}} \left( |01 \rangle + i|10 \rangle \right) & |C_{43} \rangle &= \frac{1}{\sqrt{2}} \left( |01 \rangle - i|10 \rangle \right)
\end{align*}
There are 24 possible mappings from the information basis to the complex Bell basis:
\begin{equation*}
J_{4}(w x y z) = |C_{4w} \rangle \langle 00 | + |C_{4x} \rangle \langle 01 | + |C_{4y} \rangle \langle 10 | + |C_{4z} \rangle \langle 11 |
\end{equation*}
Here \(w x y z\) is one of the permutations of the sequence 0123. These 24 possibilities fall into 5 categories. Two satisfy:
\begin{equation*}
\left[ J_{4}(2 1 3 0) \right]^{4} = \left[ J_{4}(3 0 2 1) \right]^{4} = I_{4}
\end{equation*}
Two satisfy:
\begin{equation*}
\left[ J_{4}(0 3 1 2) \right]^{8} = \left[ J_{4}(1 2 0 3) \right]^{8} = I_{4}
\end{equation*}
Two satisfy:
\begin{equation*}
\left[ J_{4}(2 1 0 3) \right]^{12} = \left[ J_{4}(3 0 1 2) \right]^{12} = I_{4}
\end{equation*}
Four satisfy:
\begin{equation*}
\left[ J_{4}(0 1 2 3) \right]^{40} = \left[ J_{4}(1 0 3 2) \right]^{40} = \left[ J_{4}(2 3 0 1) \right]^{40} = \left[ J_{4}(3 2 1 0) \right]^{40} = I_{4}
\end{equation*}
The rest satisfy
\begin{equation*}
\left[ J_{4}(w x y z) \right]^{24} = I_{4}
\end{equation*}
Let \(B_{4} \equiv J_{4}(2 1 3 0)\). Then \(B_{4}\) is a representation of \(\mathbb{Z}_{4}\). Let \(B_{8} \equiv J_{4}(0 3 1 2)\). Then \(B_{8}\) is a representation of \(\mathbb{Z}_{8}\). Let \(B_{12} \equiv J_{4}(0 2 1 3)\). Then \(B_{12}\) is a representation of \(\mathbb{Z}_{2} \times \mathbb{Z}_{6}\). Let \(B_{24} \equiv J_{4}(0 1 3 2)\). Then \(B_{24}\) is a representation of \(\mathbb{Z}_{2} \times \mathbb{Z}_{12}\). Let \(B_{40} \equiv J_{4}(0 1 2 3)\). Then \(B_{40}\) is a representation of \(\mathbb{Z}_{5} \times \mathbb{Z}_{8}\).
