Let $Q$ be a manifold, and let $X_{EL}$ be a second order vector-field on $TQ$ derived from the Euler-Lagrange equation, $$ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q} } \right) - \frac{ \partial L}{\partial q}=0,$$ for some Lagrangian $L:TQ \to \mathbb{R}$. Then for any $\alpha \neq 0$ and any $\beta \in \mathbb{R}$ we can observe that the Lagrangian $L' = \alpha L + \beta$ yields the same vector-field $X_{EL}$. So the isotropy group (acting on the space of Lagrangians) which leaves the Euler-Lagrange equatiosn unchanged appears to include $( \mathbb{R}\backslash \{0\} , \cdot) \ltimes (\mathbb{R},+)$. Is this the entire group?

If you prefer, I'd be happy to hear the answer to the same question posed on the Hamiltonian side (i.e. what is the group of transformations of Hamiltonians which leave Hamiltonian vector-fields unchanged)