I'm trying to figure out for which unit complex $z,w \in S^1$ does $zj\bar{w}=j$ for the quaternion $j$.
I was trying to solve this by setting $z=a+bi$ and $w=c+di$, so
$zj\bar{w}=(a+bi)j(c-di)=(aj+bk)(c-di)=(ac-bd)j+(bc+ad)k$
Setting this equal to $j$ we must have
(1) ac-bd=1
(2) bc+ad=0
Multiplying (1) by $b$ and (2) by $a$ and subtracting (1) from (2) we get
$b^2d-a^2d=b \implies d(b^2-a^2) = b$.
This doesn't seem to tell me anything.
Plugging $a = -\frac{bc}{d}$ into your first equation you get
$$ -\frac{bc^2}{d} - bd =1 \;\;\;\; \Longrightarrow\;\;\;\; b(c^2+d^2)= -d $$
but noting that $z$ and $w$ have unit length, this tell us that $c^2+d^2=1$ and therefore $b=-d$. Then this tell us that $a = c$. Therefore we have that $z =\overline{w}$. In other words, the transformations $\mathbb{S}^1 \times \mathbb{H} \to \mathbb{H}$ that fix $j$ are of the form
$$ (z, q) \;\; \to \;\; zqz. $$