Somebody much smarter than me told me that there is only one pair of anticommuting matrices in $\mathrm{SU}(2)$ up to conjugation. When I sat down to verify this, however, I carried out the following computation which convinced me that in fact such matrices are parameterized by $S^1$. My question then is whether they misspoke or I am making a mistake.
Suppose one is given $A,B \in \mathrm{SU}(2)$ such that $AB=-BA$. Then by appropriately conjugating we can diagonalize $B$. Suppose that after doing so, we have that $$A=\begin{bmatrix} \alpha & \beta \\ -\overline{\beta} & \overline{\alpha} \end{bmatrix}.$$ $$B=\begin{bmatrix} \gamma & 0 \\ 0 & \overline{\gamma} \end{bmatrix}$$ for some $\gamma\in S^1$ and $(\alpha, \beta) \in S^3$. Then the anticommutativity condition is precisely $$\begin{bmatrix}\alpha \gamma & -\beta \gamma \\ \overline{\beta} \overline{\gamma} & \overline{\alpha} \overline{\gamma} \end{bmatrix} = - \begin{bmatrix} \alpha \gamma & - \beta \overline{\gamma} \\ \overline{\beta} \gamma & \overline{\alpha} \overline{\gamma} \end{bmatrix}.$$ From this, we see that $\alpha=0$ and $\gamma = \pm i$. Thus, we find that, in fact $$A=\begin{bmatrix}0 & \beta \\ - \overline{\beta} & 0 \end{bmatrix}$$ and $$B=\pm\begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}.$$ This system has a $\mathbb{Z}/2$ symmetry corresponding to negation of $B$ and sending $\beta \mapsto -\overline{\beta}$. Other than this symmetry, however, I am not finding anything. What am I missing?
If $\alpha^2 = \beta$, then $$C=\begin{bmatrix}\alpha & 0 \\ 0 & \overline{\alpha} \end{bmatrix} $$ is still a $SU(2)$ matrix, and $$ C^*AC = \begin{bmatrix}0 & 1 \\ -1 & 0 \end{bmatrix} $$ $$ C^*BC = B $$