I'm trying to understand why a certain action of a Lie Group is hamiltonian.

Let $(M,g)$ be a geodesically complete Riemannian manifold. Then there exists a canonical one-form on the cotangentbundle $T^*M$ given by $\lambda_0 : T( T^*M) \to R, (v, \alpha) \mapsto \alpha(\pi_*v)$, with $\pi_*$ the derivative of the natural bundle projection $T^*M \to M$. It is now possible to pullback this one-form to the tangentbundle using our riemannian metric, so $\lambda = (g^b)^*\lambda_0$. Using this one-form we define a symplectic structure on $TM$ by $\omega = -d\lambda$.

Now let $\phi:R\times \text{TM} \to \text{TM}, V \mapsto \dot \gamma_V(t) $ be the action generated by the geodesic flow.

My question is, if $\phi$ acts by symplectomorphisms, that is, if $\phi_t^*\omega = \omega$ for all $t\in R$ or is it even exact symplectic, that is, if $\phi_t^*\lambda = \lambda$ for all $t\in R$? Because Im trying to show that this is an hamiltonian action, so that there exists a moment map for this action.

And it would be clear if we have an exact symplectic action.

Edit: To have a well-defined action of $R$, we assume that the manifold is geodesically complete.