Assuming that a Lie group $G$ acts on a smooth manifold $M$ by smooth maps (which are therefore diffeomorphisms), show that the induced action on the cotangent bundle given by $$g \cdot (x, \xi) = (g\cdot x, \varphi_{g^{-1}}^\ast\xi)$$ acts via symplectomorphisms (where $x \in M$, $\xi \in T_x^\ast M$, and $\varphi_g$ is the map which acts as $x \mapsto g \cdot x$).
Showing this amounts to showing that these actions preserve the canonical symplectic form on $T^\ast M$, but I'm not able to see why that's true.
It is actually true that for any diffeomorphism $f:M\to M$, the induced diffeomorphism $T^*f:T^*M\to T^*M$ is a symplectomorphism. This is most easily seen via the canonical one-form $\phi\in\Omega^1(T^*M)$, whose exterior derivative gives the symplectic form. By definition $\phi(\xi)(X)=\xi(T_\xi p\cdot X)$, for $X\in T_\xi (T^*M)$, where $p:T^*M\to M$ is the canonical projection (so $\xi$ is a linear map $T_{p(\xi)}M\to\mathbb R$ and thus the definition makes sense). Pulling this back along $F:=T^*f$, you get $$(F^*\phi)(\xi)(X)=\phi(F(\xi))(T_\xi F\cdot X)=F(\xi)(T_{F(\xi)}p(T_\xi F\cdot X)). $$ Now $p\circ F=f\circ p$, which shows that you get $T^*f(\xi)(T_{p(\xi)}f(T_\xi p\cdot X))$, and the definition of $T^*f$ shows that this coincides with $\phi(\xi)(X)$. Then the compatibility with the symplectic form follows from the compatibility of the exterior derivative with pullbacks.