= Solution
For $Q=\lambda q$ and $P=\lambda p$,
$$
\{Q,P\}_{q,p}=\lambda^2.
$$
The one-dimensional criterion for a <canonical transformation> therefore gives
$$
\boxed{\lambda=1\text{ or }\lambda=-1}.
$$
For either value, the <type-two generating function for a canonical transformation>
$$
\boxed{S(q,P;\lambda)=\lambda qP}
$$
works, 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.
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