[proofplan] We fix the standard complex structure $J_0$ on $\mathbb{R}^{2n}$ and observe that the subgroup of $GL_{2n}(\mathbb{R})$ commuting with $J_0$ is exactly the embedded copy of $GL_n(\mathbb{C})$. From an almost complex structure $J$ on $E$, we form the adapted frames: those real frames that intertwine $J_0$ and $J_x$. These adapted frames form a principal $GL_n(\mathbb{C})$-subbundle of $\operatorname{Fr}(E)$. Conversely, a reduced frame bundle transports $J_0$ to each fiber of $E$, and the result is independent of the chosen reduced frame because the transition functions commute with $J_0$. [/proofplan]

[step:Identify $GL_n(\mathbb{C})$ with the real automorphisms commuting with $J_0$] Let $V:=\mathbb{R}^{2n}$, and let \begin{align*} J_0:V\to V \end{align*} be the real [linear map](/page/Linear%20Map) corresponding to multiplication by $i$ under the identification $V\cong\mathbb{C}^n$. Thus $J_0^2=-\operatorname{id}_V$. Define \begin{align*} H:=\{A\in GL(V): AJ_0=J_0A\}. \end{align*} A real linear automorphism $A:V\to V$ is complex-linear for the complex [vector space](/page/Vector%20Space) structure determined by $J_0$ exactly when $A(i v)=iA(v)$ for all $v\in V$, which is exactly the condition $AJ_0=J_0A$. Hence $H$ is precisely the image of $\rho:GL_n(\mathbb{C})\hookrightarrow GL_{2n}(\mathbb{R})$. We identify $GL_n(\mathbb{C})$ with this subgroup $H$ for the rest of the proof. [/step]

[step:Construct the reduced frame bundle associated to an almost complex structure]Let \begin{align*} J:E\to E \end{align*} be a smooth bundle endomorphism over $\operatorname{id}_M$ such that $J^2=-\operatorname{id}_E$. For each $x\in M$, define \begin{align*} Q_{J,x}:=\{u:V\to E_x \text{ real linear isomorphism}: J_x\circ u=u\circ J_0\}. \end{align*} Define \begin{align*} Q_J:=\bigcup_{x\in M}Q_{J,x}\subset \operatorname{Fr}(E). \end{align*} The right action of $\operatorname{Fr}(E)$ is given by composition on the source: for $u\in \operatorname{Fr}(E)$ and $A\in GL(V)$, \begin{align*} u\cdot A:=u\circ A. \end{align*} If $u\in Q_{J,x}$ and $A\in H$, then \begin{align*} J_x\circ (u\circ A)=(J_x\circ u)\circ A=(u\circ J_0)\circ A=u\circ (J_0A)=u\circ (AJ_0)=(u\circ A)\circ J_0. \end{align*} Thus $u\cdot A\in Q_{J,x}$. Conversely, if $u\in Q_{J,x}$ and $A\in GL(V)$ satisfy $u\cdot A\in Q_{J,x}$, then \begin{align*} u\circ A\circ J_0=J_x\circ u\circ A=u\circ J_0\circ A. \end{align*} Since $u$ is injective, $AJ_0=J_0A$, so $A\in H$. Therefore the structure group preserving the adapted-frame condition is exactly $H\cong GL_n(\mathbb{C})$. It remains to note that $Q_J$ is a smooth principal $H$-subbundle. This is local on $M$. For every $x_0\in M$, choose vectors $e_1,\dots,e_n\in E_{x_0}$ such that \begin{align*} e_1,J_{x_0}e_1,\dots,e_n,J_{x_0}e_n \end{align*} is a real basis of $E_{x_0}$. Since [linear independence](/page/Linear%20Independence) is an open condition and $J$ is smooth, after shrinking to an open neighbourhood $U\subset M$ of $x_0$, there exist smooth local sections \begin{align*} s_1,\dots,s_n:U\to E \end{align*} such that \begin{align*} s_1(x),J_xs_1(x),\dots,s_n(x),J_xs_n(x) \end{align*} is a real basis of $E_x$ for every $x\in U$. These sections define a smooth local section \begin{align*} q:U\to Q_J \end{align*} by sending the standard complex basis of $\mathbb{C}^n\cong V$ to $s_1(x),\dots,s_n(x)$ complex-linearly with respect to $J_0$ on $V$ and $J_x$ on $E_x$. Therefore $Q_J$ is locally trivial with structure group $H$, hence is a $GL_n(\mathbb{C})$-reduction of $\operatorname{Fr}(E)$.[/step]

[guided]The point of the construction is to keep only those frames that make the almost complex structure on $E_x$ look like the standard one on $\mathbb{R}^{2n}$. Let $J:E\to E$ be a smooth bundle endomorphism over $\operatorname{id}_M$ with $J^2=-\operatorname{id}_E$. For each point $x\in M$, define \begin{align*} Q_{J,x}:=\{u:V\to E_x \text{ real linear isomorphism}: J_x\circ u=u\circ J_0\}. \end{align*} Thus a frame $u$ belongs to $Q_{J,x}$ precisely when it intertwines the two complex structures. Now we check the structure group. The frame bundle action is \begin{align*} u\cdot A:=u\circ A \end{align*} for $A\in GL(V)$. Suppose first that $A\in H$, so $AJ_0=J_0A$. If $u\in Q_{J,x}$, then \begin{align*} J_x\circ (u\circ A)=(J_x\circ u)\circ A=(u\circ J_0)\circ A=u\circ (J_0A)=u\circ (AJ_0)=(u\circ A)\circ J_0. \end{align*} Hence $u\circ A$ is still adapted to $J_x$. Conversely, suppose $u\in Q_{J,x}$ and $u\circ A$ is also in $Q_{J,x}$. Then the two adapted-frame identities give \begin{align*} u\circ A\circ J_0=J_x\circ u\circ A=u\circ J_0\circ A. \end{align*} Since $u:V\to E_x$ is a linear isomorphism, it is injective, so cancellation of $u$ gives $AJ_0=J_0A$. Hence $A\in H$. This proves that the allowed changes of adapted frame are exactly the complex-linear changes of basis, namely $GL_n(\mathbb{C})$. Finally we verify smooth local triviality. Fix $x_0\in M$. Since $J_{x_0}^2=-\operatorname{id}_{E_{x_0}}$, the fiber $E_{x_0}$ is a complex vector space with multiplication by $i$ given by $J_{x_0}$. Choose a complex basis $e_1,\dots,e_n$ of this complex vector space. Equivalently, \begin{align*} e_1,J_{x_0}e_1,\dots,e_n,J_{x_0}e_n \end{align*} is a real basis of $E_{x_0}$. Extend the vectors $e_1,\dots,e_n$ to smooth local sections \begin{align*} s_1,\dots,s_n:U\to E \end{align*} over some neighbourhood $U$ of $x_0$. Because linear independence of a list of $2n$ vectors is an open condition and because $J$ and the sections $s_i$ are smooth, after shrinking $U$ we may assume that \begin{align*} s_1(x),J_xs_1(x),\dots,s_n(x),J_xs_n(x) \end{align*} is a real basis of $E_x$ for every $x\in U$. Define \begin{align*} q:U\to \operatorname{Fr}(E) \end{align*} by sending the standard complex basis of $\mathbb{C}^n\cong V$ to $s_1(x),\dots,s_n(x)$ and extending complex-linearly, where the complex structure on $V$ is $J_0$ and the complex structure on $E_x$ is $J_x$. This means exactly that \begin{align*} J_x\circ q(x)=q(x)\circ J_0 \end{align*} for every $x\in U$, so $q(U)\subset Q_J$. The smoothness of $q$ follows from the smoothness of the sections $s_i$ and of $J$. These local adapted frames show that $Q_J$ is a smooth principal $H$-subbundle of $\operatorname{Fr}(E)$.[/guided]

[step:Transport the standard complex structure along a reduced frame] Conversely, let \begin{align*} Q\subset \operatorname{Fr}(E) \end{align*} be a principal $H$-subbundle. For each $x\in M$ and each $u\in Q_x$, define a real linear map \begin{align*} J_x:E_x\to E_x \end{align*} by \begin{align*} J_x:=u\circ J_0\circ u^{-1}. \end{align*} This definition is independent of the choice of $u\in Q_x$. Indeed, if $v\in Q_x$, then because $Q_x$ is an $H$-torsor, there exists a unique $A\in H$ such that $v=u\circ A$. Since $A$ commutes with $J_0$, \begin{align*} v\circ J_0\circ v^{-1}=u\circ A\circ J_0\circ A^{-1}\circ u^{-1}=u\circ J_0\circ u^{-1}. \end{align*} Thus $J_x$ is well-defined. For each $x\in M$, \begin{align*} J_x^2=(u\circ J_0\circ u^{-1})\circ (u\circ J_0\circ u^{-1})=u\circ J_0^2\circ u^{-1}=-\operatorname{id}_{E_x}. \end{align*} Hence the maps $J_x$ assemble to a fiberwise almost complex structure on $E$. [/step]

[step:Check smoothness of the transported endomorphism] We prove that the fiberwise maps $J_x$ obtained from $Q$ define a smooth bundle endomorphism \begin{align*} J:E\to E \end{align*} over $\operatorname{id}_M$. Since $Q$ is a smooth principal subbundle, every point of $M$ has an open neighbourhood $U\subset M$ and a smooth local section \begin{align*} q:U\to Q. \end{align*} Over $U$, the formula becomes \begin{align*} J_x=q(x)\circ J_0\circ q(x)^{-1}. \end{align*} In the smooth local frame $q$, the matrix of $J_x$ is the constant matrix of $J_0$. Therefore $J|_{\pi^{-1}(U)}$ is smooth. Since smoothness is local on $M$, $J:E\to E$ is a smooth bundle endomorphism satisfying $J^2=-\operatorname{id}_E$. [/step]

[step:Verify that the two constructions are inverse to each other] Start with an almost complex structure $J:E\to E$ and construct $Q_J$. If $u\in Q_{J,x}$, then \begin{align*} J_x\circ u=u\circ J_0. \end{align*} Composing on the right with $u^{-1}$ gives \begin{align*} J_x=u\circ J_0\circ u^{-1}. \end{align*} Thus transporting $J_0$ along any frame in $Q_J$ recovers the original endomorphism $J$. Conversely, start with a principal $H$-subbundle $Q\subset\operatorname{Fr}(E)$ and construct $J_x=u\circ J_0\circ u^{-1}$ for $u\in Q_x$. Then every $u\in Q_x$ satisfies \begin{align*} J_x\circ u=(u\circ J_0\circ u^{-1})\circ u=u\circ J_0. \end{align*} Hence $u\in Q_{J,x}$, so $Q_x\subset Q_{J,x}$. For the reverse inclusion, let $v\in Q_{J,x}$ and choose $u\in Q_x$. Since $u$ and $v$ are frames of $E_x$, there is a unique $A\in GL(V)$ such that $v=u\circ A$, namely $A=u^{-1}\circ v$. The adapted-frame condition for $v$ gives \begin{align*} u\circ J_0\circ u^{-1}\circ u\circ A=v\circ J_0=u\circ A\circ J_0. \end{align*} Canceling $u$ gives $J_0A=AJ_0$, so $A\in H$. Since $Q_x$ is closed under the right action of $H$, $v=u\circ A\in Q_x$. Therefore $Q_{J,x}\subset Q_x$, and hence $Q_J=Q$. The two assignments are inverse to each other. This proves the natural bijection between smooth almost complex structures on $E$ and $GL_n(\mathbb{C})$-reductions of $\operatorname{Fr}(E)$. [/step]