Since the <Hadamard gate> conjugates $X$ to $Z$,
$$
V_n=e^{i\pi(X\otimes X)/n},
\qquad
W_n=e^{i\pi(X\otimes Z)/n}.
$$
For $n=1$, each exponential is $-I$, so
$$
V_1W_1=I=e^{iA_1},
\qquad
\boxed{A_1=0}.
$$
For $n=2$,
$$
V_2=iX\otimes X,
\qquad
W_2=iX\otimes Z.
$$
Using $XZ=-iY$ gives
$$
V_2W_2=-I\otimes XZ=iI\otimes Y
=\exp\left(i\frac\pi2I\otimes Y\right),
$$
so one convenient logarithm is
$$
\boxed{A_2=\frac\pi2I\otimes Y}.
$$
Both products are <Clifford operations>: the first is the identity and the second is a one-qubit <Pauli Y gate> up to global phase.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 324 / 4 / a / iii / Solution

{c}
{title2=$Y$}
{wiki=Pauli_matrices}

The Pauli $Y$ gate is $Y=iXZ$ and maps $|0\rangle$ to $i|1\rangle$ and $|1\rangle$ to $-i|0\rangle$.


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 Incoming links: Pauli Y gate