Let $E$ be a smooth real vector bundle of even rank, over a smooth manifold $M$. Suppose there exist an orientation-reversing vector bundle isomorphism $\Phi:E \to E$.
Is it true that $E$ has a subbundle of rank $1$?
(We don't need an orientation on $E$, since for maps from a vector space to itself, the notion of orientation-preserving or reversing is always well-defined, without the need to actually choose an orientation on the space).
Here is one approach (which was suggested to me by Amitai Yuval, who also raised this question):
We can put a metric on $E$, and take the orthogonal polar factor $Q$ of $\Phi$, which will now be an isometric orientation-reversing isomorphism. So, $Q$ will have at least one non-zero fixed point at each fiber $E_x$ (see below**).
My hope is that somehow we can extract a continuously changing family of fixed points along the different fibers (one at each fiber) that will form a subbundle of rank $1$. Of course, there can be problems of multiplicity, so perhaps some perturbation argument is needed.
Can this approach work?
**Here we use the fact $\text{rank}(E)$ is even: at each fiber $Q_x$ is essentially an orthogonal matrix with negative determinant. Since its complex eigenvalues comes in conjugate pairs, and the determinant is real negative, there must be some real negative eigenvalues. Since $\det Q_x<0$ the number of the negative eigenvalues must be odd. Since $\dim E_x$ is even, we conclude there must be a positive eigenvalue, which must be $1$, since $Q_x$ is orthogonal.
Motivation:
I am trying to find out which real vector bundles admit orientation-reversing isomorphisms. Of course, every bundle of odd rank admits one: the map $x \to -x$. Now suppose the rank is even. If $E$ admits a subbundle $F$ of rank $1$, we can define an orientation-reversing isomorphism as follows:
$$\Phi|_F=\text{Id}_F,\Phi|_{{F}^{\perp}}=-\text{Id}|_{{F}^{\perp}}$$
where ${F}^{\perp}$ is some complement of $F$.
My question is if this condition ("there exist a subbundle of rank $1$") is necessary.