Suppose $X$ is an $2n$-dimensional smooth manifold with a symplectic form $\omega$. Then $\omega^n$ is a nowhere vanishing top form, so it determines an orientation on $X$. My question is: Is there a symplectic form $\omega'$ on $X$ that determines the opposite orientation?
If $n$ is odd, then clearly $-\omega$ will do this, but I can't see whether this is true if $n$ is even.
Not necessarily. For example, consider $\overline{\mathbb{CP}^2}$ which is the smooth manifold $\mathbb{CP}^2$ but equipped with the opposite orientation than the one induced by the usual complex structure. For any $[\omega] \in H^2(\overline{\mathbb{CP}^2})$, the class $[\omega]^2$ is a non-positive multiple of the class of an orientation form because $\overline{\mathbb{CP}^2}$ has intersection form $[-1]$. In particular, there can be no symplectic form which induces the given orientation.