= Solution
For a covector $p\in T_x^*X$, define the <cotangent lift of a diffeomorphism> by
$$
\boxed{f_\#(x,p)=\bigl(f(x),p\circ(df_x)^{-1}\bigr).}
$$
The inverse derivative here maps $T_{f(x)}X$ to $T_xX$. This is a smooth <diffeomorphism>, its inverse is $(f^{-1})_\#$, and it satisfies $\pi\circ f_\#=f\circ\pi$.
For $v\in T_{(x,p)}T^*X$, the <pullback of a differential form> and the defining formula for the <Liouville one-form> give
$$
\begin{aligned}
(f_\#^*\alpha)_{(x,p)}(v)
&=(p\circ(df_x)^{-1})\bigl(d\pi_{f_\#(x,p)}df_\#(v)\bigr)\\
&=(p\circ(df_x)^{-1})(df_xd\pi_{(x,p)}v)\\
&=p(d\pi_{(x,p)}v)=\alpha_{(x,p)}(v).
\end{aligned}
$$
Hence \b[$f_\#^*\alpha=\alpha$], and applying the <exterior derivative> also proves $f_\#^*\omega=\omega$. The <cotangent lift of a diffeomorphism> thus preserves both canonical forms, with no extra choice of metric or connection.
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