This is a rather naive attempt to construct an invariant of an exotic 4-sphere. Apparently, the lack of useful invariants in this context is a well known issue. This particular invariant is somewhat obvious which probably means it is useless - most likely, always zero. However, I would like to know for sure.
Consider a smooth manifold $M$ together with a homeomorphism $f: \mathbb{S}^n\to M.$ There is a natural map (the suspension) $$\phi: \mathbb{S}^{n-1}\times I\to \mathbb{S}^n$$ collapsing two $n-1$ - spheres at the ends to the poles of the $n$ - sphere, which we may interpret as a homotopy $\phi_t: \mathbb{S}^{n-1}\to \mathbb{S}^n$. For each $t\in I$ we have an oriented vector bundle $\phi_t ^* f^* TM$ on $\mathbb{S}^{n-1}$. This bundle is obviously trivial on the ends, so we may turn it into a bundle $E\to \mathbb{S}^{n-1}\times \mathbb{S}^{1}$ simply by gluing this ends.
It is easy to see that if $M$ is the standard sphere then $E$ is trivial. For an exotic sphere this depends on the homotopy class $\pi_{n-1}(SO(n))$ of the corresponding clutching function which may be trivial or not. If I am not mistaken, this construction (unlike the common one using twists) works in all dimensions, including $n=4$. In particular, it gives an invariant of an exotic 4-sphere in the group $$\pi_3(SO(4))\cong \mathbb{Z}\oplus \mathbb{Z}.$$ ([M. Kervaire. Some nonstable homotopy groups of Lie groups.(1960)].) The question is, is this invariant trivial?
