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Having answered my initial two questions (v1), I now consider a third possibility. Consider two Lagrangians that both lead to equivalent equations of motion. Suppose that they are not related via a total time derivative and can not be written in the form $L\neq L'+\frac d{dt}\phi$.

Do they lead to the same conserved quantities by Noether's theorem?

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    $\begingroup$ What is a "Noether symmetry"? Do you mean a continuous global symmetry to which you can apply Noether's first theorem? Also, how can Lagragians be "related by a gauge symmetry" - typically, the Lagrangian is invariant under a gauge symmetry? $\endgroup$ – ACuriousMind Sep 20 '15 at 13:18
  • $\begingroup$ @Qmechanic Thank you for the links, they most definitely answer the first question so I will remove that part from my question. $\endgroup$ – AngusTheMan Sep 20 '15 at 13:26
  • $\begingroup$ @ACuriousMind Yes that is what I mean, thank you. I will update my question accordingly. :) $\endgroup$ – AngusTheMan Sep 20 '15 at 13:27
  • $\begingroup$ For the existence of Lagrangians with the same e.o.m. which are not related by the same e.o.m. see this post. Same e.o.m. need not imply the same symmetries of the action/Lagrangian, though, cf. this question about how symmetries of action, e.o.m. and solutions of e.o.m. are interrelated, $\endgroup$ – ACuriousMind Sep 20 '15 at 13:49
  • $\begingroup$ Related to the question (v3): physics.stackexchange.com/q/51327/2451 and links therein. $\endgroup$ – Qmechanic Sep 20 '15 at 13:59

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