Consider the symplectic space $(\mathbb{R}^2, \omega_{st}=dx \wedge dy$. I want to find a Hamiltonian function, such that its time-1-flow is of the form
$\varphi(x,y)= (2x, \dfrac{1}{2} y)$
My ansatz: Consider $\varphi_t(x,y):=(2xt, \dfrac{1}{2} y t)$. Now
$\dfrac{d}{dt} \varphi_t(x,y)= (2x, \dfrac{1}{2} y)$
The corresponding vector field would be $X(x,y)= 2x \ dx + \dfrac{1}{2} y \ dy $.
Let $v= a \ dx + b \ dy$
Then $\omega_{st}(X,v)=2xb-\dfrac{1}{2} y a$
Is that correct so far or is it complete nonsense?
Let's look for a flow of the form $$\varphi_t(x,y) = \left(2xf(t), \dfrac{y}{2f(t)}\right)$$ for some function $f(t)$. No matter what happens, this'll always be a symplectomorphism. Since $\varphi_0 = {\rm Id}_{\Bbb R^2}$, we need $f(0) = 1/2$ and $\varphi_1= \varphi$ gives $f(1) = 1$. Moreover, the group-property of the flow now implies that $f$ is affine. So we necessarily have $f(t) = (t+1)/2$. With this we have that $$X_{(x,y)} = \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \varphi_t(x,y) = \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \left(x(t+1), \frac{y}{t+1}\right) = x\partial_x\big|_{(x,y)}-y\partial_y\big|_{(x,y)}.$$Now we have that $$\omega(X, \cdot) = ({\rm d}x\wedge {\rm d}y)(x\partial_x-y\partial_y, \cdot) = \begin{vmatrix} x & {\rm d}x \\ -y & {\rm d}y\end{vmatrix} = x\,{\rm d}y + y\,{\rm d}x = {\rm d}(xy). $$Alright, so there you have it: $H(x,y) = xy$.