Is the canonical $G$-equivariant map between two fibres of a Principal $G$-bundle always smooth? Let $\pi: P \rightarrow M$ be a principal $G$-bundle over a smooth manifold $M$. Now given any two points $p_1, p_2 \in P$ there exist a canonical function $f_{p_{1},p_2}:\bar{p_1} \rightarrow \bar{p_2}$ from  the unique fibre $\bar{p_1}$ containing $p_1$ to the unique fibre $\bar{p_2}$ containing $p_2$ which is defined as the unique $G$-equivariant function which sends $p_1$ to $p_2$ i.e $p_1.g \mapsto p_2.g$ for all $g \in G$.
My question is the following:
Is $f_{p_1,p_2}$ always smooth for any $(p_1,p_2) \in P \times P$?
 A: This canonical map is smooth, and it can be derived from a simpler one: For every $p\in P$ there is a canonical equivariant diffeomorphism $\theta_p:G\to\bar{p}$. This map is simply the restriction of the action map $\theta_p(g)=p\cdot g$. Your map is then given by $f_{p_1,p_2}=\theta_{p_2}\circ\theta_{p_1}^{-1}$.
To show that $\theta_p$ is a diffeomorphism, first note that the $G$-action restricts to a smooth, free, and transitive action on each fiber (since fibers are embedded), so $\theta_p:G\to\bar{p}$ is smooth and bijective. It now suffices to show that $\theta_g$ is an immersion (see this question). This is the case, as can be shown using a condraditiction.
Suppose $\theta_p$ is not an immersion. There must be some nonzero $v\in T_gG$ with $d_g\theta_p(v)=0$. Let $V$ be the right-invariant vector field with $V(g)=v$. Since $\theta_p$ is equivariant, $d_h\theta_p(V(h))=0$ for all $h\in G$. Let $\gamma_V:(-\epsilon,\epsilon)\to G$ be an integral curve satisfying $\gamma_V(0)=e$ and $\frac{d}{dt}\gamma_V(t)=V(\gamma_V(t))$ (one can show $\gamma_V$ extends to a Lie group homomorphism $\mathbb{R}\to G$, but that isn't needed here). We then have
$$
\frac{d}{dx}\theta_g(\gamma_V(g))=d_{\gamma_V(t)}\theta_g\left(\frac{d}{dt}\gamma_v(t)\right)=d_{\gamma_V(t)}\theta_g\left(V(\gamma_V(t))\right)=0
$$
thus $\theta_g(\gamma_V(t))=p$ and the action is not free; a contradiction.
