Take a $G$-bundle to be a fiber bundle with typical fiber $F$ whose transition functions take values in the structure group $G$ = Aut($F$). Based on this definition alone, I would assume that the tangent bundle $TM$ with typical fiber $\Bbb{R}^d$ has structure group Aut($\Bbb{R}^d$) = GL($d$, $\Bbb{R}$), and that the frame bundle $FM$ with typical fiber GL($d$, $\Bbb{R}$) has structure group Aut(GL($d$, $\Bbb{R}$)). So, the structure groups are clearly different.
However, $TM$ and $FM$ are GL($d$, $\Bbb{R}$)-associated, which entails that they share the exact same transition functions. One can see this by transforming a pointwise trivialization of $FM$:
$$\psi^B_{FM, p}([e_i]_{i=1}^d) = (\psi^B_{TM, p}(e_i))^d_{i=1} = (g^{BA}_p\psi^A_{TM,p}(e_i))_{i=1}^d = g^{BA}_p(\psi^A_{TM,p}(e_i))_{i=1}^d = g^{BA}_p\psi^A_{FM,p}([e_i]^d_{i=1})$$
To clarify the notation, we have a pointwise trivialization on $FM$ over $p \in U^X$ given by $\psi^X_{FM,p}: F_pM \to \text{GL}(d)$, a pointwise trivialization on $TM$ over $p \in U^X$ given by $\psi^X_{TM,p}: T_pM \to \Bbb{R}^d$, and a GL($d$)-valued transition function $g^{BA}_p$ transforming a local trivialization over $U^A$ to a local trivialization over $U^B$.
It seems as if the above two paragraphs contradict each other, where treating $TM$ and $FM$ as separate $G$-bundles leads to them having distinct structure groups, while treating them as GL($d$)-associated immediately implies that they have coinciding structure groups. What is my mistake here? I'm definitely missing something.
Best Answer
There seems to be an issue in your definition of a $G$-bundle. In particular, it is not the case that $G=\operatorname{Aut}(F)$. A $G$-bundle is typically defined as a collection of all of the following pieces of data:
To obtain the principal $G$-bundle associated with a $G$-bundle, we can replace the typical fiber $F$ with $G$, and the action $\theta$ with the left action of $G$ on itself. The allows us to construct a unique (up to isomorphism) bundle with the same transition functions.