 Solution

Moser's trick states that if $M$ is compact and ${\omega }_t$, $0\le t\le 1$, is a smooth family of symplectic forms whose de Rham cohomology class is independent of $t$, then there is an isotopy ${\varphi }_t$ withSince $[{\dot{\omega }}_t]=0$, choose a smooth family of one-forms ${\sigma }_t$ with ${\dot{\omega }}_t=d{\sigma }_t$. Nondegeneracy of ${\omega }_t$ uniquely determines a vector field $X_t$ byCompactness makes its flow ${\varphi }_t$ exist for the whole interval. Cartan's magic formula and $d{\omega }_t=0$ givewhich proves the theorem.

Smooth degree-$d$ hypersurfaces form the complement of the discriminant in the projective space of degree-$d$ homogeneous polynomials. This complement is path connected, so $X$ and $X^{\prime }$ lie in a smooth one-parameter family. The Ehresmann fibration theorem identifies the fibers smoothly. Under such an identification, the restrictions of the Fubini-Study form form a family ${\omega }_t$ whose cohomology class is the fixed restricted hyperplane class. Moser's trick therefore gives the symplectic equivalence of smooth projective hypersurfaces.

It remains to construct the finite subgroup for one convenient hypersurface. On the Fermat hypersurfacethe group $({\mu }_d{)}^{n+1}$ acts by diagonal coordinate multiplication. It preserves both $X_F$ and the Fubini-Study form. The kernel of its projective action is the diagonal subgroup ${\mu }_d$, so the effective Fermat hypersurface diagonal symmetry group isConjugating this action by a symplectomorphism $X_F\to X$ gives the required subgroup of $\mathrm{Symp}(X,{\omega }_{FS}{|}_X)$.