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I'm looking for an example of the following situation:

$M$ is a complex manifold, $E_1, E_2$ are non-isomorphic holomorphic vector bundles on $M$, but $E_1$ is isomorphic to $E_2$ as a complex smooth vector bundle on $M$.

Does such a thing exist, or are holomorphic structures on complex vector bundles unique?

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    $\begingroup$ See this question which asks about the case where the underlying smooth bundle is trivial. $\endgroup$
    – Michael Albanese
    Aug 30, 2022 at 10:46
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    $\begingroup$ You've got a good (+1) answer, but in case a concrete example helps, if $E$ is an elliptic curve and we fix a point $p_0$ of $E$, then for each $p$ in $E$ the divisor $[p_0] - [p]$ defines a topologically trivial holomorphic line bundle. This bundle is trivial if and only if $p = p_0$, however, since a non-trivial section of a trivial bundle is non-vanishing. $\endgroup$
    – Andrew D. Hwang
    Sep 5, 2022 at 15:35

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First, note that $\mathcal{C}^{\infty}$ is a soft sheaf, hence acyclic.

It follows that $H^1(M,(\mathcal{C}^{\infty})^{\times}) \cong H^2(M,\mathbb{Z})$ for any smooth manifold $M$. In particular, if $M$ is compact, there are only countably many isomorphism classes of smooth line bundles.

Now assume that $M$ is a compact Riemann surface of nonzero genus. Then its Jacobian $J$ parametrizes a subset of isomorphic classes of line bundles on $M$, and $J$ has uncountably many points – so some of them correspond to line bundles that are smoothly isomorphic.

(Using more carefully the ideas of First Chern class for smooth line bundle , we should be able to prove that every holomorphic line bundle of degree zero on a Riemann surface is smoothly trivial).

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