Connecting the quaternion formulation with the complex line through the hypersphere in Hopf's fibrations. I am reading about this topic mainly using this introduction by David Lyons and a presentation by Niles Johnson, and I understand several lines of approach, but I need to make the connection between the interpretation of a fiber as the result of cutting through $S^3$ with a complex line $z_2=a z_1,$ where the complex axes $z_1$ and $z_2$ are used to contain the hypersphere: $\vert z_1\vert^2 + \vert z_2\vert^2=1,$ and $a\in \mathbb C,$ and the quaternion formulation as conjugation.
In the complex line (real plane) interpretation, each $z_2= a z_1$ would define a circle (with an infinity point to be defined separately). This circle through $S^3\in \mathbb C^2$ is a fiber in the Hopf's fibration, eventually projected stereographically to $\mathbb R^3.$ The complex number $a$ can be parameterized by a point in the $S^3$ sphere in $\mathbb R^3,$ since it has $2$ degrees of freedom.
In the quaternion approach, the Hopf map can be defined as the conjugation:
$$\begin{align}
h(q)&=q \mathrm {\hat i} q^{-1} \\[2ex] 
&= (w + x\hat i + y \hat j + z \hat k)(0 + 1\hat i + 0\hat j + 0 \hat k)(w - x \hat i - y \hat j - z \hat k)\\[2ex]
&= 0 + (w^2 + x^2 - y^2 - z^2)\mathrm{\hat i} + 2 (zw + xy)\mathrm{\hat j} + 2(zx - wy)\mathrm{\hat k}
\end{align}$$
which can be thought of as a point in the $S^2$ sphere with imaginary axes. Further, if the quaternion $q$ is such that $w^2 + x^2 + y^2 + z^2=1,$ then  $$0^2 +(w^2 + x^2 -y^2 - z^2)^2 + (2(zw +xy))^2 + (2(zx - wy))^2 =1,$$ i.e. every $q \in S^3$ gets mapped to a point in $S^2.$ Or in words, conjugating any point in the unit hypersphere with one of the imaginary axes (in this case $\mathrm {\hat i}=(0,1,0,0)$) unit vectors amounts to a rotation that luckily allows the original point to be mapped to the $S^2$ sphere because the first component will always be $0.$
I don't know that there is any insight to be gained from thinking of this rotation as either a rotation around the imaginary axis $\mathrm{\hat i},$ or equivalently, of the vector $\mathrm{\hat i}$ around the axis in 4-space defined by the imaginary vector $\small\left(0, \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \frac{z}{\sqrt{x^2 + y^2 + z^2}}\right)$  by $\small \theta= 2\arctan2\left(\sqrt{(x^2 + y^2 + z^2)},w \right).$
I don't see clearly why the preimage of this quaternion interpretation necessarily has to be a circle in $\mathbb R^4,$ except for in specific examples, such as the circle in a plane in 4-space $\{(\cos t, \sin t, 0, 0)\}$ corresponding to the fiber mapping to $(0,1,0,0).$
Perhaps the way to see that this is indeed the case is by considering that rotating the unit vector $\small (0, 1 \mathrm {\hat i},0,0)$ around the axis of a purely imaginary vector $\small\left(0, \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \frac{z}{\sqrt{x^2 + y^2 + z^2}}\right)$  from $0$ to $2\pi$ will define a circle, and this can be parameterized by $w.$ Given that this rotating of $\mathrm {\hat i}$ around and axis and angle determined by the quaternion $(w,x,y,z)$  is exactly how the Hopf's map is defined, and that there is an equivalence between the quaternion and its axis-angle representation, it makes sense that the preimage of a point in $S^2$ is that set of quaternions that rotate $\mathrm {\hat i}$ around the same imaginary axis to form a full circle. However, I can't make it happen using the interactive presentation by Ben Eater, allowing conversion from the trigonometric to the quaternion forms, and trying to rotate the $\mathrm {\hat i}$ unit vector around the unit vector $0.78 \mathrm {\hat i} + 0 \mathrm {\hat j} + 0.63 \mathrm {\hat k}$ at first by $2\theta = 2 \times 17^o =34^o,$ and later by $2\theta=2 \times 70^o=140^o,$ corresponding to the quaternions $0.96 + 0.23 \mathrm {\hat i} + 0 \mathrm {\hat i} + 0.18 \mathrm {\hat k}$ and $0.34 + 0.73 \mathrm {\hat i} + 0 \mathrm {\hat i} + 0.59 \mathrm {\hat k},$ which wouldn't map to the same point - instead, they would map to $(0,0.9421,0.3456,0.0828)$ and $(0,0.3004,0.4.12,0.8614),$ respectively.

If this did work, I could think that the axis marked by this imaginary vector forms a line in $\mathbb C^2$ through the origin, which links to the slope $a.$ But it doesn't...
The orbit stabilizer of the action of a set of the $3$-sphere on the vector $(0,1,0,0)$ forms a plane in $4$-space. For other points in $\mathbb S^2,$ [not comfortable at all with this, but I found this very useful] the stabilizer or pre-image will also be a unit circle, but I guess I can't just choose any unitary quaternion and conjugate with $\mathbf i,$ like in the illustration above, and assume that the 'revolution' of $\mathbf i$ around the new vector will map to the same point: the circles stabilizing other points are linked to the $\mathbf i$ stabilizer: they are cosets of $\mathbb S^1$ formed by $q(\cos(\theta)+i\sin(\theta))$ where $q \in \mathbb S^3.$
QUESTION:

How can I convince myself that all Hopf fibers are circles in $\mathbb R^4$ by connecting the complex line idea and quaternions?

 A: To rotate a quaternion in $\mathbb{R}^4$, consider the basis matrices:
The left Cayley matrices:
\begin{equation}
\hat{l}_i\;=\; 
\begin{pmatrix}
0 & -1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0
\end{pmatrix}\;\;\;\;
\hat{l}_j\;=\; 
\begin{pmatrix}
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0
\end{pmatrix}\;\;\;\;
\hat{l}_k\;=\; 
\begin{pmatrix}
0 & 0 & 0 & -1 \\
0 & 0 & -1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0
\end{pmatrix}
\end{equation}
which satisfy the relation
$$\hat{l}_i^2\;=\;\hat{l}_j^2\;=\;\hat{l}_k^2\;=\;\hat{l}_i\hat{l}_j\hat{l}_k\;=\;-\hat{l}_1$$
The right Cayley matrices:
\begin{equation}
\hat{r}_i\;=\; 
\begin{pmatrix}
0 & 1 & 0 & 0 \\
-1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0
\end{pmatrix}\;\;\;\;
\hat{r}_j\;=\; 
\begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0
\end{pmatrix}\;\;\;\;
\hat{r}_k\;=\; 
\begin{pmatrix}
0 & 0 & 0 & 1 \\
0 & 0 & -1 & 0 \\
0 & 1 & 0 & 0 \\
-1 & 0 & 0 & 0
\end{pmatrix}
\end{equation}
which satisfy the relation
$$\hat{r}_i^2\;=\;\hat{r}_j^2\;=\;\hat{r}_k^2\;=\;\hat{r}_i\hat{r}_j\hat{r}_k\;=\;-\hat{r}_1$$
and $\hat{l}_1,\hat{r}_1$ are the $4\times4$ identity matrices.
In vector form the quaternion is represented
\begin{equation*}
\vec{\Psi}\;=\;
\begin{pmatrix}
q_1 \\
q_i \\
q_j \\
q_k 
\end{pmatrix} 
\end{equation*}
Rotate the quaternion $\vec{\Psi}$ in the left Cayley basis, through an angle $\varphi$ in the $\hat{l}_i$ axis.
$$\vec{\Psi}'\;=\;\exp\left(\frac{\varphi}{2}\hat{l}_i\right)\vec{\Psi}$$
\begin{equation}
\vec{\Psi}'\;=\;
\begin{pmatrix}
\cos\left(\tfrac{\varphi}{2}\right) & -\sin\left(\tfrac{\varphi}{2}\right) & 0 & 0 \\ 
\sin\left(\tfrac{\varphi}{2}\right) & \cos\left(\tfrac{\varphi}{2}\right) & 0 & 0 \\ 
0 & 0 & \cos\left(\tfrac{\varphi}{2}\right) & -\sin\left(\tfrac{\varphi}{2}\right) \\
0 & 0 & \sin\left(\tfrac{\varphi}{2}\right) & \cos\left(\tfrac{\varphi}{2}\right) 
\end{pmatrix}
\begin{pmatrix}
q_1 \\
q_i \\
q_j \\
q_k 
\end{pmatrix}
\;=\; 
\begin{pmatrix}
 \cos\left(\tfrac{\varphi}{2}\right)q_1-q_i\sin\left(\tfrac{\varphi}{2}\right) \\ 
 \cos\left(\tfrac{\varphi}{2}\right)q_i+q_1\sin\left(\tfrac{\varphi}{2}\right) \\ 
 \cos\left(\tfrac{\varphi}{2}\right)q_j-q_k\sin\left(\tfrac{\varphi}{2}\right) \\ 
 \cos\left(\tfrac{\varphi}{2}\right)q_k+q_j\sin\left(\tfrac{\varphi}{2}\right)
\end{pmatrix}
\end{equation}
By extension - the quaternion can be rotated in a circle in $\mathbb{R}^4$ using the 6 rotation axes $\hat{l}_i,\hat{l}_j,\hat{l}_k,\hat{r}_i,\hat{r}_j,\hat{r}_k$.
The stereographic projection of the quaternion, is the map
\begin{equation*}
 \mathbb{S}^3/(1,0,0,0)\;\mapsto\;\mathbb{R}^3
\end{equation*}
given by
\begin{equation} 
 (q_1',q_i',q_j',q_k')\;\mapsto\;\left(\frac{q_i'}{1-q_1'},\frac{q_j'}{1-q_1'},\frac{q_k'}{1-q_1'}\right)
\end{equation}
with $\varphi\in[0,4\pi]$.
