Isomorphism between the principal $S^1$-bundles $\rho_*(S^3)$ and $SO(3)$ Let $\rho:H\rightarrow G$ be a Lie group homomorphism and suppose that $\pi:P\rightarrow M$ is a principal $H$-bundle. Then, one can define a principal $G$-bundle $\rho_*(P)$ over $M$, which is given by the quotient of $P\times G$ modulo the $H$-action
$$
(p,g)h=(ph,g\rho(h)).$$
Each element in $\rho_*(P)$ is of the form $[p,g]$ and there is a "projection" $\widetilde{\pi}:\rho_*(P)\rightarrow M$ such that $\widetilde{\pi}([p,g])=\pi(p)$. Furthermore, the $G$-action is given by
$$ [p,g]g'=[p,gg'].$$
With that set, consider the principal $S^1$-bundles $S^3$ and $SO(3)$ (over $S^2$). We are perceiving $S^1$ as the set of unitary complex elements and $(z_0,z_1)\in S^3\subset \mathbb{C}^2$, with $|z_0|^2+|z_1|^2=1$. Also, the former bundle is completely characterized by the projection $\pi:S^3\rightarrow S^2$ and action by $\lambda\in S^1$:
\begin{align*}
\pi(z_0,z_1)&=(|z_0|^2-|z_1|^2,2i\overline{z_0}z_1),\\
(z_0,z_1)\lambda&=(z_0\lambda,z_1\lambda).
\end{align*}
Similarly, for $A=(A^1,A^2,A^3)\in SO(3)$ (with $A^i$ denoting the $i$-th column) and $\lambda=e^{i\alpha}$ for some $\alpha\in\mathbb{R}$, we have $\tilde{\pi}:SO(3)\rightarrow S^2$ given by
\begin{align*}
&\tilde{\pi}(A^1,A^2,A^3)=A^1,\\
&A\cdot\lambda=A\begin{pmatrix}1&0&0 \\ 0&\cos(\alpha)&\sin(\alpha)\\0&-\sin(\alpha)&\cos(\alpha)\end{pmatrix}.
\end{align*}
Consider then $\rho:S^1\rightarrow S^1$ given by $\rho(z)=z^2$. I would like to prove that $\rho_*(S^3)\simeq SO(3)$, as principal bundles. Now, I know there is a function $\phi:S^3\rightarrow SO(3)$ which takes $u\in S^3$ (viewed as an unit quaternion) to the function
$$\phi(u)(v)=u^*vu,$$
for every $v\in\mathbb{R}^3$ and I already know that this function is smooth and satisfies $\tilde{\pi}\circ\phi=\pi$ and $\phi(u\lambda)=\phi(u)\cdot\lambda^2$. The idea is to use $\phi$ to define a function $\tilde{\phi}:\rho_*(S^3)\rightarrow SO(3)$ which commutes with the projections and is $S^1$-equivariant.
I tried putting $\tilde{\phi}([u,\lambda])=\phi(u)\cdot\lambda^{-1}$, which satisfied everything I needed except being $S^1$-equivariant, so maybe I am wrong and in fact it is equivariant, or maybe there is another choice for $\tilde{\phi}$. Any helps are appreciated!
 A: @Jason DeVito's comment shed some light on the problem and I believe there is a mistake in the set of notes I am reading. I believe the action of $H$ over $P\times G$ should read
$$ (p,g)h=(ph,\rho(h)^{-1}g).$$
In that case, we must define $$\widetilde{\phi}([u,\lambda])=\phi(u)\cdot\lambda.$$ To make it more of an answer, I will provide the proof that, indeed, this is the required isomorphism between $\rho_*(S^3)$ and $SO(3)$. We must have

*

*$\widetilde\phi$ is well-defined,

*$\widetilde{\pi}\circ\widetilde\phi=\pi$,

*$\widetilde\phi$ must be $S^1$-equivariant.

So let's do this:
1)) If $[u,\lambda]=[u',\lambda']$, then there is $\widetilde\lambda\in S^1$ such that
$$(u',\lambda')=(u\widetilde\lambda,\widetilde\lambda^{-2}\lambda'),$$
so putting everything into $\widetilde\phi$ and using the fact that $\phi(u\lambda)=\phi\cdot\lambda^2$, we have
\begin{align*}
\widetilde\phi([u',\lambda'])&=\phi(u')\cdot\lambda'\\
&=\phi(u\widetilde\lambda)\cdot(\widetilde\lambda^{-2}\lambda)\\
&=\phi(u)\cdot\lambda\\
&=\widetilde\phi([u,\lambda]).
\end{align*}
2)) This is straight-forward from the definition of $\widetilde\phi$ and invariance of $\widetilde\pi$ with respect to the $S^1$ action.
3)) Take $\widetilde\lambda\in S^1$ and compute
\begin{align*}
\widetilde\phi([u,\lambda]\widetilde\lambda])&=\widetilde\phi([u,\lambda\widetilde\lambda])\\
&=\phi(u)\cdot(\lambda\widetilde\lambda)\\
&=\widetilde\phi([u,\lambda])\cdot\widetilde\lambda,
\end{align*}
as wanted.
