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Edit: I misunderstood the question...

The answer to the first question for closed orientable manifolds in dimension 4 is negative.

The Dold-Whitney theorem states that two oriented 4-plane bundles over the same 4-manifold $M$ are isomorphic if and only if they share the same second Stiefel-Whitney class $w_2$, the same first Pontryagin class $p_1$, and the same Euler class $e$. Each such characteristic class is determined by the homotopy type of $M$ (see below), and hence the cotangent bundle over $M$ is also determined, no matter what differentiable structure one assigns to $M$.

These characteristic classes are indeed easily determined by the homotopy type of $M$: $p_1$ is 3 times the signature $\sigma(M)$ of $M$ by Hirzebruch formula, the Euler class is $\chi(M)$, and $w_2$ is determined by the intersection form thanks to Wu's formula, see for instance this question.

2 stupid typo...

The answer to the first question for closed orientable manifolds in dimension 4 is positivenegative.

The Dold-Whitney theorem states that two oriented 4-plane bundles over the same 4-manifold $M$ are isomorphic if and only if they share the same second Stiefel-Whitney class $w_2$, the same first Pontryagin class $p_1$, and the same Euler class $e$. Each such characteristic class is determined by the homotopy type of $M$ (see below), and hence the cotangent bundle over $M$ is also determined, no matter what differentiable structure one assigns to $M$.

These characteristic classes are indeed easily determined by the homotopy type of $M$: $p_1$ is 3 times the signature $\sigma(M)$ of $M$ by Hirzebruch formula, the Euler class is $\chi(M)$, and $w_2$ is determined by the intersection form thanks to Wu's formula, see for instance this question.

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The answer to the first question for closed orientable manifolds in dimension 4 is positive.

The Dold-Whitney theorem states that two oriented 4-plane bundles over the same 4-manifold $M$ are isomorphic if and only if they share the same second Stiefel-Whitney class $w_2$, the same first Pontryagin class $p_1$, and the same Euler class $e$. Each such characteristic class is determined by the homotopy type of $M$ (see below), and hence the cotangent bundle over $M$ is also determined, no matter what differentiable structure one assigns to $M$.

These characteristic classes are indeed easily determined by the homotopy type of $M$: $p_1$ is 3 times the signature $\sigma(M)$ of $M$ by Hirzebruch formula, the Euler class is $\chi(M)$, and $w_2$ is determined by the intersection form thanks to Wu's formula, see for instance this question.