Edit: I think my original formula for the induced action had some mistakes, which I believe I've corrected.
If a Lie group $G$ acts on a manifold $X$, there is an induced action on $T^*X$ by pulling back. Explicitly, if $\alpha=(p,\alpha_p) \in T^*X$, we can define a new one-form $g\cdot \alpha \in T^* X$ above $g \cdot p$: for any $v=(g\cdot p,v_{g\cdot p}) \in TX$ $$ (g\cdot\alpha)(v):= \alpha_p\big((\rho_{g^{-1}})_*(v_{g\cdot p})\big) = \alpha_p\big((\rho_g^{-1})_*(v_{g\cdot p})\big)\equiv \big((\rho_g^{-1})^*\alpha_p\big)(v_{g\cdot p}) $$ where $\rho_g: X \to X$ is the diffeomorphism of $g$ acting on $X$. $T^*X$ is canonically an exact symplectic manifold with primitive one-form given by, for $v = (p,v_p) \in T(T^*M)$ with $p \in T^*M$ and $v_p \in T_p(T^*M)$, $$ \lambda(p,v_p) := p \cdot \pi_*(v_p) $$ and symplectic form given by $\omega = d\lambda$. In coordinates, $T^*X$ has the form $(q_1,\dots,q_n,p_1,\dots,p_n)$ where $q_i$ are coordinates on the base space and $p_i$ are the fiber coordinates. We have $$ \lambda = \sum_i p_i dq_i $$ We must be careful with notation here since we are dealing with an induced action by pullbacks. If $\rho_g: X \to X$ is the diffeomorphism associated to the action of $g$ on $X$, then $(\rho_g^{-1})^*: T^*X \to T^*X$ is the induced diffeomorphism on cotangent bundle. Denote this by $\Phi_g$. Then we must show that $$ \big((\rho_g^{-1})^*\big)^*\omega \equiv(\Phi_g)^*\omega = \omega $$ Note that $\Phi_g^*$ preserves $\omega$ if it preserves $\lambda$, by naturality of pullbacks wrt differentials. So it suffices to compute $$ \Phi_g^*(\lambda) = \Phi_g^*\left(\sum_i p_idq_i\right) $$ $$ = \sum_i \Phi_g^*p_i \wedge d\left(\Phi_g^*q_i\right) $$ $p_i$ and $q_i$ are functions on the manifold $T^*U$, so pullback acts by precomposition: $$ =\sum_i (p_i \circ \Phi_g)d\left(q_i \circ \Phi_g\right) $$
But I do not know how to proceed from here. I have encountered some coordinate free calculations online but right now I'm interested computing the pullback in coordinates directly.