I'm lacking intuition and robust mathematical understanding of this. Let $(LM,\pi,M)$ be a frame bundle and let us define a right $GL(d,\mathbb R)$-action $\triangleleft$ on $LM$, s.t $\triangleleft:LM\times GL(d,\mathbb R)\to LM$, where: $$(e_1,\cdots,e_d)\triangleleft g:=({g^m}_1e_m,\cdots,{g^m}_de_m)$$
We know that $(e_1,\cdots,e_d)$ is a frame chosen from $LM$, where $LM=\underset{x\in M}{\dot{\bigcup}}L_xM$ as a set and $L_xM$ is the set of all $(e_1,\cdots,e_d)$ that constitute a basis for the tangent space of the manifold $M$ at the point $x$. With $d$ we denote the dimension of $M$ and $g$ belongs to the $GL$ group and can be thought of as an invertible general linear transformation matrix with matrix elements ${g^m}_n$.
My first question is about the freedom of this right action. Somehow, I have to prove that for every element $p$ of $LM$, the stabiliser $S_p$ is only the identity on $GL(d,\mathbb R)$, i.e only the right action of $e$ leaves the frame invariant. I really cannot think of a counterexample to this, the new frame vectors will always be linear combinations of the initial frame vectors, so for both frames to be equal, $g=e$ (considering that $(0,\cdots,0)$ cannot be a basis for the tangent space). My attempt to prove this formally goes like this: $$e'_i={g^m}_ie_m$$ where the i-th row is fixed and we sum over the columns of the transformation matrix. Suppose $g\neq e$ and $e'_i=e_i$, i.e every frame vector is invariant under an arbitrary GL transformation. Then $$e_i={g^m}_ie_m\qquad i=1,\cdots,d\qquad m=1,\cdots,i,\cdots,d$$ so that $${g^m}_ie_m-e_i={g^i}_ie_i-e_i+{g^k}_ie_k=({g^i}_i-1)e_i+{g^k}_ie_k=0\qquad k=1,\cdots,\not{i},d$$ but because $\{e_i\}$ is a basis, all vectors must be linearly independent, i.e the above equation holds only if the coefficients ${g^\bullet}_\bullet$ are zero. As a result, for all $i$, all ${g^k}_i$ must be zero and all ${g^i}_i$ must equal 1. This is the identity, so case proven. Do you find this proof acceptable?
We want this to be a principal $GL(d,\mathbb R)$-bundle, so the last step is to prove that $(LM,\pi,M)$ is isomorphic as a bundle to $(LM,\rho,LM/_{GL(d,\mathbb R)})$, where the base space $LM/_{GL(d,\mathbb R)}$ is the quotient space $LM/_\sim$, i.e for $q\in LM$, $$LM/_\sim:=\{p\in LM\mid p\sim q:\iff \exists\:g\in GL(d,\mathbb R):p=q\triangleleft g\}$$ or the way I understand this, $[q]\in LM/_\sim$ is the equivalence class of all $q\in LM$ that have the same orbit $\mathcal O_q$. The projection map $\rho$ simply maps $q$ to $[q]$.
What I need to show is that there exist diffeos $$u:LM\to LM \qquad f:M\to LM/_{GL(d,\mathbb R)}$$ ($u,f$ obviously smooth and invertible), that satisfy the following equality: $$\rho\circ u= f\circ \pi $$ I'm terribly confused at this point, so the best thing I can do is share my intuition so that you can correct me.
If I have an element $l$ of $LM$ and I act on it from the right with the whole GL group I should get all possible orbits of $l$. I imagine it as a set of all $\mathcal O_l$ sets, so if I pick an element out of it, it would be a new frame, say $j$, still an element of $LM$. Now, $\rho$ maps this $j$ to $[j]\in LM/_{GL(d,\mathbb R)}$, i.e all those elements of $LM$ that have the same orbit as $j$. Where is the structure preservation if on one base space we have points on $M$ and on the other we have frames on $M$? How should $f$ look like? I mean, I think of $LM$ as points and frames on those points. I understand that if we mod out the frames, then we are left with the points, these frames were residing on, but how exactly do $LM$ modulo $GL$ and equivalence classes mean the same?
I'm sorry for the long text. I'm a physics student, new to principal bundles, so please be as patient as you can (there might be many mistakes above). I would like someone to guide me especially through the last part and if possible propose a math formulation for the proof. Thank you.