Diffeomorphic but not isotopic symplectic forms Do we know of any closed symplectic manifold $M$ with 2 cohomologous symplectic forms $\omega_1$ and $\omega_2$ such that there exist $\psi \in \text{Diff}(M)$ and $\psi^* \omega_1 = \omega_2$ but $\omega_1$ and $\omega_2$ are not isotopic?
Here the word isotopy is supposed to mean that the two forms a joined by a path of cohomologous symplectic forms. 
 A: This answer is an extension of my last comment, which in turn is just a reference to this answer by MO user Petya, and the paper Symplectic Topology and Capacities by McDuff cited therein.
First, your condition for two forms to be isotopic is equivalent to the existence of a path of diffeomorphisms $\phi_t$ such that $\phi_{0} = \operatorname{id},\phi_1^*\omega_2 = \omega_1$. One direction is clear; in the other, let $\omega_t$ be the family of symplectic forms and define a time-dependent vector field $X_t$ by $\mathrm d\left(\iota_{X_t}\omega_t\right) = \partial_t\omega_t$ (by assumption, the RHS is exact). The flow of $X_t$, which exists by compactness, is the required family of diffeomorphisms.
Take $M = S^2\times S^2\times T^2$, with $T^2 = S^1\times S^1$ the torus, and let $\omega_1 = p_1\omega_{S^2} + p_2^*\omega_{S^2} + p_3^*\omega_{T^2}$ where $\omega_{S^2}$ and $\omega_{T^2}$ are volume forms on $S^2$ and $T^2$, respectively. (Note that we use the same volume form for both factors of $S^2$.) Let $\psi(x,y,u,v) = (x,R(x,v)y,u,v)$, where $R:S^2\times S^1\to SO(3)$ sends $(x,v)$ to the rotation by $v$ around the axis through $x$. For fixed $y_0,u_0$, the map $x,v\mapsto (x,y_0,u_0,v)$ defines an $S^1$-family of $J$-holomorphic spheres for the product complex structure, and the projection of this family to the second $S^2$-factor si constant. Since each of these spheres has minimal positive area, there is no bubbling, so that for any deformation one obtains a compact cobordism between this family and its pushforward under the end of the deformation. In particular, the projection of the pushforward to the second factor is bordant to a constant map. But one can check that the map $f: S^2\times S^1\to S^2,(x,v)\mapsto R(x,v)y_0$ is not bordant to a constant map: Its restriction to $S^2\vee S^1$ is nullhomotopic, and the resulting map $S^2\times S^1/(S^2\vee S^1)\cong S^3\to S^2$ is homotopic to the Hopf map, which is not nullbordant.
