I am trying to understand the following theorem about symplectomorphisms of projective bundles. Theorem 1.5 of "Characteristic Classes in Symplectic Topology" A.G. Reznikov. Selecta Mathematica, volume 3, pages 601–642(1997).

Theorem: Let $E_i \rightarrow M_i$ be Hermitian vector bundles, $i = 1, 2$. Let $ \mathbb{P}(E_i)$ be the projectivization of $E_i$. Let $f : \mathbb{P}( E_1) → \mathbb{P} (E_2)$ be a fiber-like symplectomorphism, covering a map $\varphi : M_1 \rightarrow M_2$. Then $\varphi^{*}(c_k(E_2)) = c_k(E_1)$ for $k \geq 2$.

Here is where I get confused: taking a simple case where $M_1=M_2=\,\mathbb{CP}^2$. Suppose that $E_{1} = \mathcal{O} \oplus \mathcal{O}$. $E_{2} = \mathcal{O}(1) \oplus \mathcal{O}(1)$. Then, if I am not mistaken, there should be a fibre-wise symplectomorphism $\mathbb{P}(E_1) \rightarrow \mathbb{P}(E_2)$ since they are both isomorphic to the $3$-fold $\mathbb{CP}^2 \times \mathbb{CP}^1$. But there is no diffeomorphism of $\mathbb{CP}^2$ mapping $c_{2}(\mathcal{O} \oplus \mathcal{O}) = 0$ to $c_{2}(\mathcal{O}(1) \oplus \mathcal{O}(1)) = H^2$, where $H$ is the hyperplane class (it is not hard to see that a fibre preserving symplectomorphism must induce a diffeomorphism of the base).

I am thinking possibly that this theorem is for a specific choice of symplectic form?

I would be grateful if somebody could clear up my confusion. I am by no means questioning this theorem, I just want to understand the statement correctly. thanks.