In this question, it is shown that:
$$L=\frac{1}{2}(v^2_1+v_2^2-q_1^2-q_2^2)$$ and $$H=\frac{1}{2}(p^2_1+p_2^2+q_1^2+q_2^2)$$ both have a symetry group $U(2)$ despite their form that would suggest $O(2,2)$ and $O(4)$. Is the following affirmation true:
The symmetries $O(4)$ and $O(2,2)$ are off-shell symmetry of the action since they hold without the equation of motion and thus are the one for which Noether theorem applies. Which implies that, when talking about the Noether theorem in this case, the (dynamical?) $U(2)$ symmetry never shows up. We can conclude that two different symmetries implies the same conservation law.
