Let $H:\mathbb{R}^{2n}\rightarrow \mathbb{R}$ be a Hamiltonian function, $\phi:\mathbb{R}^{2n}\rightarrow \mathbb{R}^{2n}$ be a symplectic transformation, that is $(D_x\phi(x)) J (D_x\phi(x))^T = J$ for $J=\left(\begin{array}{cc} 0 & -I \\ I & 0 \\ \end{array}\right)\in\mathbb{R}^{2n\times 2n}$, $C\in\mathbb{R}^{n\times 2n}$ be a linear transformation satisfying $CJC^T=0$ and $K :\mathbb{R}^{n}\rightarrow \mathbb{R}$ be a function.
Question: For what Hamiltonians function $H(x)$ do there exist a $\phi$, $C$ and $K$ satisfying $$ H(\phi(x)) = K(C \phi(x))$$
Example: If the Hamiltonian is a function of $p$ only H(p), then $\phi = id$, $C = (I,0)$ and $K=H$ will satisfy the above. The same could be said about functions of $q$ only $H(q)$. This also holds for when you can write $H(x) = K_1(Cx)$ for $CJC^T=0$. I would like to know more general classes of Hamiltonians for which there does not exist such a $K_1$ satisfying $H(x) = K_1(Cx)$, but there does exist one after a symplectic change of variables $ H(\phi(x)) = K(C \phi(x))$. We do not need to know what $\phi$, $K$ or $C$ are, just need to know that they exist.
Some background on why we choose this condition for $C$ can be found here.