https://en.wikipedia.org/wiki/Hopf_fibration
What is the group of transformations $\subset SO(4)$ that sends every fibre circle to another fibre circle?
I think the Lie algebra might be generated by the bivectors
$$(e_1e_2),(e_3e_4),(e_1e_3+e_2e_4),(e_1e_4+e_3e_2)$$
or, in terms of matrices,
$$\begin{bmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix},\begin{bmatrix}0&0&0&0\\0&0&0&0\\0&0&0&-1\\0&0&1&0\end{bmatrix},\begin{bmatrix}0&0&-1&0\\0&0&0&-1\\1&0&0&0\\0&1&0&0\end{bmatrix},\begin{bmatrix}0&0&0&-1\\0&0&1&0\\0&-1&0&0\\1&0&0&0\end{bmatrix}$$
but this is just a guess. And I don't know much about Lie algebras, or what the corresponding Lie group would be.
...Is it the intersection of $SO(4)$ with the symplectic group $Sp(4,\mathbb R)$ ? Does that have a name?
This was solved by Gluck, Warner, and Ziller in
(https://www.researchgate.net/publication/266548925_The_geometry_of_the_Hopf_fibrations).
Here's a summary:
The symmetry group of the Hopf fibration $S^{2n+1}\rightarrow \mathbb{C}P^n$ is $U(n+1) \cup c U(n+1)$ where $c:\mathbb{C}^{n+1}\rightarrow \mathbb{C}^{n+1}$ is conjugation: $c(z_1,.., z_{n+1}) = (\overline{z}_1,..., \overline{z}_{n+1})$. For "the" Hopf fibration $S^3\rightarrow S^2$, the symmetry group is $U(2)\cup cU(2)\subseteq SO(4)$.
The symmetry group of the Hopf fibration $S^{4n + 3}\rightarrow \mathbb{H}P^n$ is $Sp(n+1)\times S^3/\langle (-I, -1)\rangle$.
The symmetry group of the Hopf fibration $S^{15}\rightarrow S^8$ is $Spin(9)$ acting by its spin representation.