Fibre bundle with a free and proper right-action that is isomorphic to the principal fibre bundle from the Quotient Manifold Theorem. Fibres = Orbits?

Question

Suppose $\pi\colon P \to M$ is a fibre bundle, $(G,\bullet)$ a Lie group and $\triangleleft\colon P \times G \to P$ a (smooth) free and proper Lie group right action of $G$ on $P$. The Quotient Manifold Theorem then states that $\rho\colon P \to P/{\triangleleft G}, a \mapsto \operatorname{Orb}_{\triangleleft}(a) = a \triangleleft G$ is a principal bundle.

My doubt is the following: I originally suspected that if $\triangleleft$ preserves the fibres of $\pi\colon P \to M$ and if there exists a bundle isomorphism $\require{AMScd}\varphi\colon P \to P$ along $f\colon M \to P/{\triangleleft G}$, then $\pi\colon P \to M$ is a principal bundle with the property that its fibres are precisely given by the group orbits.

Now, however, I find myself unable to prove that there exists a diffeomorphism $\tilde f\colon M \to P/{\triangleleft G}$ such that $\mathrm{id}_P\colon P \to P$ is a bundle isomorphism along $\tilde f\colon M \to P/{\triangleleft G}$.

I was intending to argue in the following way:

1. I can define a smooth map $$F\colon P/{\triangleleft G} \to P/{\triangleleft G}, \quad a \triangleleft G \mapsto (\rho \circ \varphi)(a),$$ which is well-defined since $\triangleleft$ preserves fibres and since $\rho \circ \varphi = f \circ \pi$. This map then satisfies $F \circ \rho = \rho \circ \varphi$.
2. Define a map $$\tilde F\colon P/{\triangleleft G} \to P/{\triangleleft G}, \quad a \triangleleft G \mapsto (\rho \circ \varphi^{-1})(a)$$ and prove that it is well-defined, smooth and the inverse to $F$.
3. Then $\rho = F^{-1} \circ f \circ \pi$ follows, thus constructing a diffeomorphism $F^{-1} \circ f\colon M \to P/{\triangleleft G}$ such that $\mathrm{id}_P\colon P \to P$ is a bundle isomorphism along $F^{-1} \circ f\colon M \to P/{\triangleleft G}$.

In order to prove well-definedness in Step 2, it looks like I would have to require from the beginning that the fibres of $\pi\colon P \to M$ are precisely given by the group orbits. Yet, intuitively it feels to me like this assumption shouldn't be necessary.

Is my intuition actually valid? If so, how do I prove the statement? Or does there exist a counterexample?

Edit 1: Topological Complications?

Let us look at the following modification of the question above:

Suppose $\pi\colon P \to M$ is a fibre bundle with a (smooth) free and proper Lie group right action $\triangleleft\colon P \times G \to P$ of $G$ on $P$ such that $\triangleleft$ preserves the fibres of $\pi\colon P \to M$. Is it possible that there exists a bundle isomorphism $\varphi\colon P \to P$ along $f\colon M \to P/{\triangleleft G}$ while there does not exist a diffeomorphism $\tilde f\colon M \to P/{\triangleleft G}$ such that $\mathrm{id}_P\colon P \to P$ is a bundle isomorphism along $\tilde f$?

Example: (smooth) free and proper $U(1)$-action on two disjoint cylinders as total space

Consider the following example. The total space $P$ shall be given as the disjoint union of two cyliners with the base space $M = (-1,1)$ being an open interval. Suppose we have a fibre-preserving, free and proper right-action of $U(1)$ on the total space. Then the orbit space $P/{\triangleleft G}$ has 2 connected components while the base space $M$ is connected. Therefore no bundle isomorphism between $\rho\colon P \to P/{\triangleleft G}$ and $\pi\colon P \to M$ can exist. Refer to the sketch below: Does this argument hold true in the general case as well, stating that there cannot exist a bundle isomorphism whatsoever, if at least one fibre is given as the disjoint union of at least two orbits?

Answer 1

I think your intuition is valid. I believe you can define an inverse of your function $f:M\to P/{\triangleleft G}$ as follows: define $h:P/{\triangleleft G}\to M$ by $h(a\triangleleft G) := \pi(\varphi^{-1}(a))$. $h$ is well-defined since the group action preserves fibres of $\pi:P\to M$, as does $\varphi^{-1}$. It is smooth, since locally it can be written as $h\vert_U = \pi\circ\varphi^{-1}\circ s$, where $s:U\to P$ is some local section of $\rho:P\to P/{\triangleleft G}$. It can be checked that $f\circ h = \mathrm{id}_{P/{\triangleleft G}}$ and $h\circ f =\mathrm{id}_M$. So $f$ is a diffeomorphism.

Edit: if I'm understanding the example in your edit correctly, it seems to me that $P/{\triangleleft G}$ would consist of two copies of $(-1,1)$ (one for each cylinder). You say this, but it's not reflected in your picture. Then I'm not sure what you mean by "a bundle isomorphism between $\rho:P\to P/{\triangleleft G}$ and $\pi:P\to M$" in this context, since a bundle isomorphism of principal bundles will only exist if the fibres are isomorphic, which is not true in your case. Unless your definition of principal bundle isomorphism is different to mine?

Edit 2: we have by assumption \begin{CD} P @>{\varphi}>> P \\ @VV{\pi}V @VV{\rho}V\\ M @>{f}>>P/{\triangleleft G} \end{CD} Assuming you accept the fact that $f$ has an inverse $h = f^{-1}$ (I'm still not clear whether you do), $f$ is a diffeomorphism. Since $\varphi$ is a bundle isomorphism, we also have \begin{CD} P @>{\varphi}>> P \\ @VV{\pi}V @VV{\pi}V\\ M @>{\check{\varphi}}>>M \end{CD} for some diffeomorphism $\check{\varphi}$. Combining these two diagrams gives \begin{CD} P @>{\mathrm{id}_P}>> P \\ @VV{\pi}V @VV{\rho}V\\ M @>{f\circ\check{\varphi}^{-1}}>>P/{\triangleleft G} \end{CD} and $f\circ\check{\varphi}^{-1}$ is a diffeomorphism. So it acts as your searched for $\tilde{f}$.