Suppose we have a ball of mass $m$ in the Earth's gravitational field ($g=const.$). Equation of motion reads as: $$ ma = -mg $$ From here we can conclude that we have translational symmetry of the form $x(t) \to x(t) + const$ (we are working in only 1D). However, we cannot see this symmetry from the Lagrangian: $$ L = \frac{mv^2}{2} - mgx $$ because the linear term "breaks" this symmetry. Moreover, we also do not have the corresponding conserved quantity (as far as I can see).
Does this mean that we can have symmetries in the Newtonian sense (transformations that map solutions to other solutions) that are not present in the Lagrangian?