Let $\mathbb{R}^{2n}\times\mathbb{R}=\mathbb{R}^n\times\mathbb{R}^n\times\mathbb{R}=\{(q,p,z)|{q},{p}\in\mathbb{R}^n,z\in\mathbb{R}\}$ be a contact manifold with the standard contact form $\alpha=pdq+dz$. Then we have the $\mathbb{R}_{>0}$-quotient of its symplectization $Y$ where $Y=T^*\mathbb{R}^n\times T^*_{>0}\mathbb{R}=\{(q,p,z,\zeta)|(q,p)\in T^*\mathbb{R}^n,(z,\zeta)\in T^*\mathbb{R},\zeta>0\}$ with natural $\mathbb{R}_{>0}$-action gien by $F_\lambda(q,p,z,\zeta)=(q,\lambda p,z,\lambda\zeta)$.
For given a contact isotopy $\{\Phi_s\}_{s\in I}:\mathbb{R}^n\times\mathbb{R}\rightarrow\mathbb{R}^n\times\mathbb{R}$, in the Chiu's paper "Non-squeezing property of contact balls", he said that this isotopy can be lifted to the conic Hamiltonian symplectic isotopy, by using the same notation, $\{\Phi_s\}_{s\in I}:Y\rightarrow Y$ satisfying a conic $\mathbb{Z}$-equivariant condition $\Phi_s\circ F_\lambda=F_\lambda\circ\Phi$.
So, if the given contact isotopy is $\Phi_s(q,p,z)=:(f_s(q,p,z),g_s(q,p,z),h_s(q,p,z))$, I defined this lift by $Phi_s(q,p,z,\zeta)=(f_s(q,p/\zeta,z),\zeta g_s(q,p/\zeta,z),h_s(q,p/\zeta,z),\zeta)$. Then the conic $\mathbb{Z}$-equivariant condition holds.
However, I cannot check this isotopy is symplectic. Is the definition wrong? If it is, how to construct the lift?
