Let us consider a mechanical system with coordinates $q^k$ ($k = 1, 2, \dots, N$) and a Lagrangian $L(q, \dot{q}, t)$. We introduce infinitesimal transformations of time and trajectories \begin{equation} t \to \tilde{t} = t + \epsilon^\alpha v_\alpha(t) + \mathcal{O}(\epsilon^2), \quad q^k(t) \to \tilde{q}^k(\tilde{t}) = q^k(t) + \epsilon^\alpha V^k_\alpha(q(t), t) + \mathcal{O}(\epsilon^2), \end{equation} with $M$ parameters $\epsilon^\alpha$, ($\alpha = 1, 2, \dots, M$).
Note that the global form of these transformations can be written as \begin{equation} \tilde{t} = f(t, \epsilon), \quad \tilde{q}^k(\tilde{t}) = F^k(q(t), t, \epsilon), \end{equation} where the map from $t$ to $\tilde{t}$ is assumed invertible, i.e. $t = f^{-1}(\tilde{t}, \epsilon)$. If We have a symmetry transformation, meaning that \begin{equation} d\tilde{t} \, L(\tilde{q}(\tilde{t}), \dot{\tilde{q}}(\tilde{t}), \tilde{t}) = dt \, L(q(t), \dot{q}(t), t) + dB(q, t). \end{equation} and $q(t)$ is a solution of the Euler-Lagrange equations, then \begin{equation} \tilde{q}^k(t) = F^k(q(f^{-1}(t, \epsilon)), f^{-1}(t, \epsilon), \epsilon) \end{equation} also satisfies the equations of motion, In specific cases I checked this to be true, but I can't prove it in general. I do understand that Euler-Lagrange equations are satisfied w.r.t transformed time and transformed generalized coordinates.
