[proofplan] We prove both directions by comparing the same bundle vector in two overlapping trivializations. If a global bundle isomorphism exists, its local matrix representatives must commute with transition maps, and this gives the displayed cocycle conjugacy relation. Conversely, if the conjugacy relation holds, the local formulas $(x,v)\mapsto (x,h_i(x)v)$ are compatible on overlaps, so they glue to a global smooth bundle map. Since each $h_i(x)$ is invertible, the glued map has a smooth inverse obtained from the local matrices $h_i(x)^{-1}$. [/proofplan]

[step:Extract the cocycle relation from a global bundle isomorphism]Assume first that $\Phi:E \to E'$ is a smooth vector bundle isomorphism over $\operatorname{id}_M$. For each $i \in I$, let \begin{align*} h_i:U_i \to GL(k,\mathbb{R}) \end{align*} be the smooth local representative of $\Phi$ in the chosen trivializations, meaning that in $U_i \times \mathbb{R}^k$ coordinates, \begin{align*} (x,v) \mapsto (x,h_i(x)v). \end{align*} Fix $i,j \in I$ and $x \in U_i \cap U_j$. Let $v \in \mathbb{R}^k$ denote an arbitrary vector written in the $j$-trivialization of $E$. Changing first from the $j$-trivialization to the $i$-trivialization in $E$ sends $v$ to $g_{ij}(x)v$, and then applying $\Phi$ in the $i$-trivialization sends it to \begin{align*} h_i(x)g_{ij}(x)v. \end{align*} Applying $\Phi$ first in the $j$-trivialization sends $v$ to $h_j(x)v$, and then changing from the $j$-trivialization to the $i$-trivialization in $E'$ sends it to \begin{align*} g'_{ij}(x)h_j(x)v. \end{align*} Both procedures describe the same vector of the fiber $E'_x$ in the $i$-trivialization, so \begin{align*} h_i(x)g_{ij}(x)v=g'_{ij}(x)h_j(x)v \end{align*} for every $v \in \mathbb{R}^k$. Hence \begin{align*} h_i(x)g_{ij}(x)=g'_{ij}(x)h_j(x). \end{align*} Since $h_j(x) \in GL(k,\mathbb{R})$, right multiplication by $h_j(x)^{-1}$ gives \begin{align*} g'_{ij}(x)=h_i(x)g_{ij}(x)h_j(x)^{-1}. \end{align*}[/step]

[guided]Assume that a global smooth vector bundle isomorphism \begin{align*} \Phi:E \to E' \end{align*} over $\operatorname{id}_M$ is already given. Because $\Phi$ is fiberwise linear and covers the identity on $M$, its expression in the trivialization over $U_i$ has the form \begin{align*} (x,v) \mapsto (x,h_i(x)v), \end{align*} where \begin{align*} h_i:U_i \to GL(k,\mathbb{R}) \end{align*} is smooth. The codomain is $GL(k,\mathbb{R})$ rather than all $k \times k$ matrices because $\Phi$ is a fiberwise linear isomorphism. Now fix $i,j \in I$ and $x \in U_i \cap U_j$. We compare two ways of describing the image under $\Phi$ of a vector whose $j$-coordinates in $E_x$ are $v \in \mathbb{R}^k$. First change coordinates in $E$ from the $j$-trivialization to the $i$-trivialization. By the transition convention, the vector becomes $g_{ij}(x)v$. Then apply $\Phi$ using the $i$-local formula. The resulting $i$-coordinates in $E'_x$ are \begin{align*} h_i(x)g_{ij}(x)v. \end{align*} Second apply $\Phi$ before changing coordinates. In the $j$-trivialization, the image has coordinates $h_j(x)v$. Then change coordinates in $E'$ from the $j$-trivialization to the $i$-trivialization. By the transition convention for $E'$, the resulting $i$-coordinates are \begin{align*} g'_{ij}(x)h_j(x)v. \end{align*} Let $e \in E_x$ denote the bundle vector whose coordinates in the $j$-trivialization of $E$ are $v$. These are two coordinate descriptions, in the same $i$-trivialization of $E'$, of the same vector $\Phi(e) \in E'_x$. Therefore \begin{align*} h_i(x)g_{ij}(x)v=g'_{ij}(x)h_j(x)v \end{align*} for every $v \in \mathbb{R}^k$. Since two linear maps $\mathbb{R}^k \to \mathbb{R}^k$ agreeing on every vector are equal, we obtain \begin{align*} h_i(x)g_{ij}(x)=g'_{ij}(x)h_j(x). \end{align*} Finally, $h_j(x)$ is invertible because $h_j(x) \in GL(k,\mathbb{R})$, so right multiplication by $h_j(x)^{-1}$ gives \begin{align*} g'_{ij}(x)=h_i(x)g_{ij}(x)h_j(x)^{-1}. \end{align*}[/guided]

[step:Use the cocycle relation to glue the local bundle maps] Conversely, suppose that smooth maps \begin{align*} h_i:U_i \to GL(k,\mathbb{R}) \end{align*} satisfy \begin{align*} g'_{ij}(x)=h_i(x)g_{ij}(x)h_j(x)^{-1} \end{align*} for every $i,j \in I$ and every $x \in U_i \cap U_j$. For each $i \in I$, define a smooth local bundle map \begin{align*} \Phi_i:U_i \times \mathbb{R}^k \to U_i \times \mathbb{R}^k \end{align*} by the rule that, for each $(x,v) \in U_i \times \mathbb{R}^k$, its value is \begin{align*} \Phi_i(x,v)=(x,h_i(x)v). \end{align*} We verify compatibility on overlaps. Fix $x \in U_i \cap U_j$ and $v \in \mathbb{R}^k$, where $v$ is written in the $j$-trivialization of $E$. In the $i$-trivialization of $E$, the same vector has coordinates $g_{ij}(x)v$. Applying $\Phi_i$ gives \begin{align*} h_i(x)g_{ij}(x)v. \end{align*} On the other hand, applying $\Phi_j$ first gives $h_j(x)v$ in the $j$-trivialization of $E'$, and changing to the $i$-trivialization of $E'$ gives \begin{align*} g'_{ij}(x)h_j(x)v. \end{align*} Using the assumed relation, \begin{align*} g'_{ij}(x)h_j(x)v = h_i(x)g_{ij}(x)h_j(x)^{-1}h_j(x)v = h_i(x)g_{ij}(x)v. \end{align*} Thus the local formulas agree after the transition identifications on every overlap. By the gluing principle for smooth maps defined in local trivializations, compatible smooth local bundle maps over a common open cover determine a unique smooth global bundle map. Therefore the maps $(\Phi_i)_{i \in I}$ assemble to a well-defined smooth vector bundle map \begin{align*} \Phi:E \to E' \end{align*} over $\operatorname{id}_M$. [/step]

[step:Construct the inverse from the inverse local matrices] For each $i \in I$, define \begin{align*} \Psi_i:U_i \times \mathbb{R}^k \to U_i \times \mathbb{R}^k \end{align*} by the rule that, for each $(x,w) \in U_i \times \mathbb{R}^k$, its value is \begin{align*} \Psi_i(x,w)=(x,h_i(x)^{-1}w). \end{align*} The inversion map on $GL(k,\mathbb{R})$ is smooth, so each map $x \mapsto h_i(x)^{-1}$ is smooth, and hence each $\Psi_i$ is smooth. The same overlap computation, with the equality \begin{align*} g_{ij}(x)=h_i(x)^{-1}g'_{ij}(x)h_j(x), \end{align*} shows that the maps $(\Psi_i)_{i \in I}$ assemble to a smooth bundle map \begin{align*} \Psi:E' \to E \end{align*} over $\operatorname{id}_M$. In the local trivialization over $U_i$, \begin{align*} (\Psi_i \circ \Phi_i)(x,v) = \Psi_i(x,h_i(x)v) = (x,h_i(x)^{-1}h_i(x)v) = (x,v), \end{align*} and \begin{align*} (\Phi_i \circ \Psi_i)(x,w) = \Phi_i(x,h_i(x)^{-1}w) = (x,h_i(x)h_i(x)^{-1}w) = (x,w). \end{align*} Thus $\Psi \circ \Phi=\operatorname{id}_E$ and $\Phi \circ \Psi=\operatorname{id}_{E'}$. Therefore $\Phi$ is a smooth vector bundle isomorphism represented locally by the maps $h_i$, completing the proof. [/step]