= Solution
First observe that the <Liouville one-form> is zero as an ambient covector exactly on the <zero section>. Indeed, $\alpha_{(x,p)}=p\circ d\pi$, and $d\pi$ is surjective. Since $dg$ is invertible, $g^*\alpha=\alpha$ therefore implies that $g$ maps the <zero section> onto itself. It induces a <diffeomorphism> $f:X\to X$ defined by $g(x,0)=(f(x),0)$.
The map $g$ is a <symplectomorphism>, since it preserves $-d\alpha$. Let $E=\sum_i p_i\partial_{p_i}$ be the vertical <Liouville vector field>. With the chosen sign convention,
$$
\iota_E\omega=-\alpha.
$$
Preservation of both $\omega$ and $\alpha$ forces $g_*E=E$ because $\omega$ is <nondegenerate>. Its complete <cotangent fiber-dilation flow> is
$$
D_s(x,p)=(x,e^sp),\qquad s\in\mathbb R.
$$
Uniqueness of integral curves gives $g\circ D_s=D_s\circ g$. If $g(x,p)=(y,P)$, let $s\to-\infty$. Continuity and this commutation yield
$$
(f(x),0)=g(x,0)=\lim_{s\to-\infty}g(x,e^sp)
=\lim_{s\to-\infty}(y,e^sP)=(y,0).
$$
Thus $y=f(x)$ for every $p$: the whole cotangent fiber over $x$ maps into the fiber over $f(x)$. This step proves that $g$ covers $f$; it was not assumed.
Now write $g(x,p)=(f(x),P(x,p))$. For any $u\in T_xX$, choose a tangent vector to $T^*X$ projecting to $u$. Evaluating $g^*\alpha=\alpha$ on it gives
$$
P(x,p)(df_xu)=p(u).
$$
Since $df_x$ is invertible, $P(x,p)=p\circ(df_x)^{-1}$. Therefore
$$
\boxed{g=f_\#,\qquad f=\pi\circ g\circ i_0,}
$$
and $f$ is unique. This <rigidity of cotangent Liouville-form preservation> uses smoothness at the <zero section>; it follows from the canonical form and its dilation dynamics on the entire <cotangent bundle>.
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