Any smooth manifold $X$ admits a Riemannian metric $g$, and we have a map $$ TX \to T^*X, \qquad (x, v) \mapsto (x, g(v,-)) $$ which is smooth if $g$ is. Why isn't this an isomorphism of vector bundles? We have a smooth, linear map inducing an isomorphism on all fibers; shouldn't this be enough?
For instance, on $X = \mathbb{CP}^1$, we have $TX \cong O(-2)$ and $T^*X \cong O(2)$, which are not isomorphic as evidenced by their Chern classes. Maybe I just don't understand the notion of an isomorphism of vector bundles.
A Riemannian metric does indeed give an isomorphism between $TX$ and $T^*X$ as real vector bundles. However, if $X$ is a complex manifold, then $TX$ can also be viewed as a complex vector bundle, but the isomorphism $TX\cong T^*X$ provided by a Riemannian metric is not complex linear. If the metric is Hermitian, then the isomorphism complex anti-linear. Since Hermitian metrics always exist on complex manifolds, it's always the case that $T^*X$ is complex-linearly isomorphic to $\overline {TX}$.