I think this is a well knowing result but I can't find any reference, 


Let $(E,q)$ be a vector bundle with a non degenerated quadratic form $q:E\rightarrow E^\*$ with trivial determinant, suppose that $q$ is symmetric (resp. skew-symmetric). 


My question: Is there an equivalent of the stack of such couples with the one of $(\mathbb Z/2, SL\_r)$-bundle, with the action of $\mathbb Z/2$ on $SL\_r$ is giving by $$g\rightarrow ^tg^{-1} \;(resp.\,\,g\rightarrow J\_n\;^tg^{-1}J\_n)$$ 


where $J\_n=\begin{pmatrix}0&I\_n\\-I\_n&0\end{pmatrix}$


and how can one prove that (if the answer was yes!)?


thanks.



Vector bundle with a perfect pairing and ($\mathbb Z/2$, $SL_r$)-bundle
Let me answer my last question (last comment), if we give a $\rho:G\rightarrow G$ an automorphism of $G$, let us denote by $$E^\rho=E\times^GG$$ where the action of $G$ on itself is given by $\rho$, then generally $E^\rho$ is not isomorphique to $E$. It is when $\rho$ is an interior automorphism. For exemple taking $\rho(g)=^tg^{-1}$, $G=GL\_r$, then one can see easily $$E^\rho(\mathbb C^r)\cong E(\mathbb C^r)^\*$$


