This MO post is a motivation to ask:
Is paralellizability a topological invariant (Invariant under homeomorphism)?
2026-03-25 19:11:46.1774465906
Is paralellizability a topological invariant (Invariant under homemorphism)
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No. You can cook up counterexamples in high dimensions via surgery theory. The relevant paper here is Milnor, "Microbundles and differentiable structures", available here. The theorem you want is Corollary 6.4; as a corollary of this, Milnor gives an example of a (noncompact) smooth parallelizable manifold that carries a different, non-parallelizable differentiable structure.
You should be able to find closed examples using this corollary, but I don't know any off the top of my head.