= Solution
Apply preservation of <Poisson brackets> to the coordinate functions. It gives
$$
\{Q_i,Q_j\}=0,\qquad\{P_i,P_j\}=0,\qquad\{Q_i,P_j\}=\delta_{ij}.
$$
Conversely, the entries of $AJA^T$ are exactly the <Poisson brackets> between all transformed coordinate functions. The displayed relations therefore say $AJA^T=J$, which implies the <canonical transformation> condition and, by the <chain rule>, preservation of every <Poisson bracket>. Thus \b[all three conditions are equivalent]:
$$
\boxed{f^*\omega_0=\omega_0\quad\Longleftrightarrow\quad
\text{all brackets are preserved}\quad\Longleftrightarrow\quad
\{Q_i,Q_j\}=\{P_i,P_j\}=0,\ \{Q_i,P_j\}=\delta_{ij}.}
$$
In particular the original coordinate bracket $\{q_i,p_j\}$ is $\delta_{ij}$ with the chosen <Poisson bracket> convention.
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