Q: A question about dual map on modules Let $R$ be a commutative ring (noetherian if needed), and $P$ a finitely generated projective $R$-module (of constant rank $n$). Let $P^\vee:={\rm Hom}_R(P, R)$ be the dual. For $\varphi\in{\rm Aut}_R(P\oplus R)$, define $\varphi^*, \varphi^\vee\in{\rm Aut}_R(P^\vee\oplus R)$ by letting (where $\langle -, -\rangle$ is the natural pairing between a module and its dual)
$$\langle \varphi^*(\xi), x\rangle:=\langle \xi, \varphi(x)\rangle, x\in P\oplus R, \xi\in P^\vee\oplus R$$
and $\varphi^\vee:=(\varphi^*)^{-1}$. One checkes easily that
$$\langle \varphi^\vee(\xi), \varphi(x)\rangle=\langle \xi, x\rangle, x\in P\oplus R, \xi\in P^\vee\oplus R.$$
My question is: 

For $x\in P\oplus R, \xi\in P^\vee\oplus R$ with $\langle \xi, x\rangle=1$, can we find $\varphi\in{\rm Aut}_R(P\oplus R)$ such that $\varphi(0, 1)=x, \varphi^\vee(0, 1)=\xi$? And if $\det P\cong R$, can we further get such a $\varphi$ preserving $\det P$ (i.e. $\det \varphi={\rm id}$)?

The following fact may be helpful:
Let $V:=\ker(P\oplus R\xrightarrow{\langle \xi, -\rangle}R)\subset P\oplus R$ and $V':=\ker(P^\vee\oplus R\xrightarrow{\langle -, x\rangle}R)\subset P^\vee\oplus R$, then it's easy to see that we have decompositions $P\oplus R=V\oplus Rx, P^\vee\oplus R=V'\oplus R\xi$. I want to show that $V\cong P$ but I can't (this is of course true if $P$ is a cancellative projective module).
I don't even know how to prove or give a counterexample when $P=R^n$. Any idea or proof (maybe in some special cases) or counterexample is welcome!
