 Solution

For $Q=\lambda q$ and $P=\lambda p$,The one-dimensional criterion for a canonical transformation therefore givesFor either value, the type-two generating function for a canonical transformationworks, because $p=S_q=\lambda P$ and $Q=S_P=\lambda q$; when ${\lambda }^2=1$, the first equation is equivalent to $P=\lambda p$. For every other $\lambda$, the Poisson bracket is not one and the map is not canonical.