Suppose that $(M_{1},\omega_{1})$ and $(M_{2},\omega_{2})$ are compact symplectic $4$-manifolds, that are (oriented) diffeomorphic. Is it true that the Todd genus ($\frac{1}{12} (c_{1}^{2} + c_{2})(M_{i})$) are equal?
I know a reference that the answer is yes for algebraic surfaces (since the Todd genus equals $\chi(\mathcal{O}_{S}) = 1 - \frac{1}{2}b_{1}(S) + P_{1}(S) $ and the plurigenera are known to be an (oriented) smooth invariant by Seiberg-Witten theory).
In dimension 4, the Todd genus does not depend on the choice of a symplectic structure or even on an almost complex structure. If $M$ is an almost complex 4-manifold, then $\langle c_1(M)^2, [M]\rangle = 2\chi(M) + 3\sigma(M)$ (see here, p. 9), and $\langle c_2(M), [M]\rangle = \chi(M)$; here $\chi(M)$ is the Euler characteristic of $M$ and $\sigma(M)$ is its signature.
Thus any orientation-preserving diffeomorphism preserves the Todd genus, and the orientation-preserving assumption is necessary, because the signature depends on the choice of orientation.
This is also mentioned at the end of §2 of this paper by Łukasz Bąk.
