Let $F:M\times G\rightarrow P$ be an isomorphism and $e$ the neutral element of $G$. For every $x\in E$, write $\sigma(x)=F(x,e)$. The map $\sigma$ is a section of $\pi:P\rightarrow M$ since $F$ is a morphism of bundle.
To $\sigma$ is associated the isomorphism $F_\sigma: M\times G\rightarrow P$ defined by
$F_\sigma(x,g)=\sigma(x).g=F(x,e).g=F(x,g)$.
Hey Thiago:D I really like your question; I'll illustrate my thoughts in some detail so my apologies if I say lots of really obvious things.
Fix a scheme $X$, a sheaf of groups $G$ on $X$ and a $G$-torsor $P$; suppose we work in the Zariski topology for concreteness (as in your case with $\text{GL}_n$ torsors anyway, although I'm rather sure this generally works) and let $Y = \bigcup_i U_i$ be an open cover of $Y\subset X$ such that $P_{\mid U_i}$ is isomorphic to the trivial left $G$-torsor $G_{\mid U_i} \xrightarrow{\phi_i} P_{\mid U_i}$. I first think it's worth noting how $\text{Aut}_G(P)$ is a sheaf of $G$-sets: given any automorphism of $G$-torsors $\alpha : P_{\mid Y} \to P_{\mid Y}$, we have that $\phi_i\alpha_{\mid U_i}\phi_i^{-1}: G_{\mid U_i}\to G_{\mid U_i}$ is given by right multiplication by some element $a_i \in G(U_i)$ and it's rather tempting, at least for me, to define the action on $\alpha \in \text{Aut}_G(P)(Y)$ by an element $g\in G(Y)$ in imposing $\phi_i(g\cdot \alpha)_{\mid U_i}\phi_i^{-1}$ to be right multiplication by $a_i\cdot g^{-1}$, but of course there are gluing problems... we can examine them: if we set $U_{i,j}:= U_i\cap U_j$ then we have a commutative diagram $$
\require{AMScd}
\begin{CD}
G_{\mid U_{i,j}} @>{\cdot (a_i)_{\mid U_{i,j}}}>> G_{\mid U_{i,j}}\\
@V{(\phi_i)_{\mid U_{i,j}}}VV @VV{(\phi_i)_{\mid U_{i,j}}}V \\
P_{\mid U_{i,j}} @>{\alpha_{\mid U_{i,j}}}>> P_{\mid U_{i,j}}\\
@A{(\phi_j)_{\mid U_{i,j}}}AA @AA{(\phi_j)_{\mid U_{i,j}}}A \\
G_{\mid U_{i,j}} @>{\cdot (a_j)_{\mid U_{i,j}}}>> G_{\mid U_{i,j}}
\end{CD}
$$ which reads $$
a_{i} = (\phi_j^{-1}(\phi_i(1_G))\cdot a_{j}\cdot \phi_i^{-1}(\phi_j(1_G))
$$ and thus substituting $a_i,a_j$ for $a_i\cdot g^{-1},a_j\cdot g^{-1}$ might potentially hinder this equality, preventing us from getting a new automorphism $g\cdot \alpha : P_{\mid Y} \to P_{\mid Y}$. If instead we try replacing $a_i$ and $a_j$ with $a_i^g,a_j^g$ (where I use the group theorist's notation $x^y := yxy^{-1}$) we get an equality $$
a_i^g = (\phi_j^{-1}(\phi_i(1_G))^g\cdot a_{j}^g \cdot (\phi_i^{-1}(\phi_j(1_G)))^g
$$ and the sweet thing is that we can replace the trivialisation $\phi_i$ with $\phi_i\circ(\cdot g)$ (which of course still provides an isomorphism between $G_{\mid U_i}$ and $P_{\mid U_i}$) thus defining an action of $G(Y)$ on $\text{Aut}_G(P)$ swapping $a_i$ for $a_i^g$, by the equation above. In particular, the automorphism sheaf of the trivial $G$-torsor $\text{Aut}_G(G) \cong G$ is endowed with the action by conjugation.
If we now consider the presheaf $$
F : X_{\text{Zar}}^{\text{op}} \to (\text{sets})
$$ given by $$
F(U) = P(U) \times G(U) / (h\cdot p,g)\sim (p,hgh^{-1})
$$ whose sheafification is the contracted product $F^\# = P \wedge^G G$ where $G$ acts on itself via inner automorphisms, we can construct a morphism $F \to \text{Aut}_G(P)$ by mapping the pair $(p,g)$ to the automorphism $$
P_{\mid U} \xrightarrow{\phi_p^{-1}} G_{\mid U} \xrightarrow{\cdot g} G_{\mid U} \xrightarrow{\phi_p} P_{\mid U}
$$ where $\phi_p : G_{\mid U} \to P_{\mid U}$ is the trivialisation corresponding to the section $p \in P(U)$. Then applying $(-)^{\#}$ yields an isomorphism of sheaves $P\wedge^G G \xrightarrow{\cong} \text{Aut}_G(P)$ since for the trivial torsor, as mentioned above, we have $\text{Aut}_G(G) \cong G \cong G\wedge^G G$.
Finally, turning to your particular situation, we can actually make the same considerations for $\text{Aut}_{\mathcal{O}_X}(E)$ where $E$ is a vector bundle on $X$ (by which I mean a locally free sheaf of $\mathcal{O}_X$-modules of rank $n$): the $\text{GL}_{n,X}$-torsor $P$ corresponding to $E$ is the sheaf $$
U \mapsto \text{Isom}_{\mathcal{O}_X}(E_{\mid U}, \mathcal{O}_U^{\oplus n})
$$ and thus we can define a morphism of presheaves $$
F \to \text{Aut}_{\mathcal{O}_X}(E)
$$ by mapping the pair $(p,g) \in P(U) \times \text{GL}_n(\mathcal{O}_U)$ to $$
E_{\mid U} \xrightarrow{p^{-1}} \mathcal{O}_U^{\oplus n} \xrightarrow{g} \mathcal{O}_U^{\oplus n} \xrightarrow{p} E_{\mid U}
$$ which once again defines an isomorphism $F^\# = P\wedge^{\text{GL}_n}\text{GL}_n \cong \text{Aut}_{\mathcal{O}_X}(E)$ for the same reason.
So in conclusion we have $$
\text{Aut}_{\text{GL}_{n,X}}(P) \cong P\wedge^{\text{GL}_{n,X}}\text{GL}_{n,X}
\cong \text{Aut}_{\mathcal{O}_X}(E).$$
I do want to mention that generally speaking I dislike working with cocycles... I'd love to see an explanation of the $G$ action on $\text{Aut}_G(P)$ without needing to fix a trivialisation :P
Hope this helps and I didn't mess something silly up (:
Best Answer
If $X$ is homotopy equivalent to a paracompact space, then $k_G(X) = [X, BG]$ where $BG$ is the classifying space of the topological group $G$ and the brackets denote the collection of homotopy classes of maps $X \to BG$. If $f : Y \to X$ is a continuous map, then there is an induced map $[X, BG] \to [Y, BG]$ given by $\alpha \mapsto \alpha\circ f$. If $Y$ is homotopy equivalent to a paracompact space, we can view this as a map $k_G(X) \to k_G(Y)$ which, by construction of the identification $k_G(X) = [X, BG]$, is precisely $f^*$, i.e. the map $k_G(X) \to k_G(Y)$ is given by $[P] \mapsto [f^*P]$. Now, if $f$ is a homotopy equivalence, then $f^*$ is a bijection (if $g$ is a homotopy inverse of $f$, then $g^*$ is an inverse of $f^*$).
In conclusion, if $X$ and $Y$ are homotopy equivalent to paracompact spaces, and $f : Y \to X$ is a homotopy equivalence, then $f^* : k_G(X) \to k_G(Y)$ given by $[P] \mapsto [f^*P]$ is a bijection.