In the ICTP lectures of Y. Grossman: Standard Model 1, in about minute 54:00, he leaves an informal homework for the students. He ask to find the symmetry related to the conservation of the amplitude of the simple classical harmonic oscillator.
I thought about it and it is a bit obvious from Newton's equations. A harmonic oscillator is defined (with unit constants) as
$$\ddot{x}=-x$$
and the solution is $x(t)=A\cos(t+\phi)$, where $A$ is a constant amplitude and $\phi$ a phase.
Guess 1
Replacing the solution in the equation for the energy gives
$$E=\frac{1}{2}\dot{x}^2+\frac{1}{2}x^2=\frac{1}{2}A^2$$
thus the amplitude is conserved because it is proportional to the energy, which is conserved (no dissipative forces).
Thus as a first guess, the symmetry is just the same as the symmetry for the energy conservation: time translation symmetry.
Guess 2
The equation for the amplitude is
$$A^2=2E=x^2+\dot{x}^2$$
which is the equation for a circle. Any rotation in phase space, will leave this quantity the same. So I can try the following infinitesimal transformation $\delta x = \dot{x}\epsilon$ and $\delta \dot{x}=-x\epsilon$, write the Lagrangian and input all this in the machinery of Noether's theorem.
In the end, I found that $A^2$ is conserved. This is not a surprise because I started from the equation of a circle. (This can also be verified using inverse Noether theorem to confirm that it is the good infinitesimal transformation).
Thus the symmetry is the invariance by rotations in phase space.
Which is it?
The first guess seems trivial and it is probably what it was meant but I am unable to confirm. The second guess is a bit of an artifact because I basically started from the equation of Hamiltonian times a constant, but the symmetry is a bit different that what I was expecting from guess 1.
What is the symmetry related to that question? Is there another way to independently define the amplitude which makes its connection to the Hamiltonian less straightforward?
I just watched again the part where Grossman asks the question, and he says more precisely "what is the symmetry that guarantees that the period does not depend on the amplitude?". I suppose that it is related to what I have already written here. A much more naive argument would be to say that the period $\omega^{-1}$ has to depend on $k$ (spring constant), $m$ (mass), and $A$, but by dimensional analysis $k$ and $m$ suffice $\omega\propto\sqrt{k/m}$, so what is the symmetry there?
Best Answer
The most clean derivation is to go to the Hamiltonian formulation. Then the conserved charge is the Hamiltonian $$H~=~\frac{p^2}{2m}+\frac{kq^2}{2}\tag{A}$$ (basically the square of the amplitude $A$), and it generates the infinitesimal symmetry transformation in phase space $$\begin{align}\delta q~=~&\epsilon \{q ,H\}, \cr \delta p~=~&\epsilon \{p ,H\}, \cr \delta t~=~&0. \end{align}\tag{B} $$ This is essentially OP's 2nd guess.
However, the symmetry transformation in Noether's theorem is not unique, cf. e.g. my Phys.SE answer here. For energy conservation (OP's 1st guess), this is e.g. demonstrated in my Phys.SE answer here.
Let us go to angle-action variables $$\begin{align} \varphi~:=~&\arg(p+im\omega q)~\sim~\varphi+2\pi, \cr J~:=~&\frac{1}{2\pi}\oint p \mathrm{d}q. \cr \{\varphi,J\}~=~&1. \end{align}\tag{C}$$ Then the SHO Hamiltonian (A) becomes $$H~=~\omega J,\quad\text{where}\quad \omega~:=~\sqrt{k/m}.\tag{D}$$
The fact that the period doesn't depend on the amplitude is encoded by the constant of motion $$Q~=~\varphi -\omega t.\tag{E}$$ It generates the infinitesimal quasi-symmetry transformation
$$\begin{align} \delta \varphi~=~&\epsilon \{\varphi ,Q\}~=~0, \cr \delta J~=~&\epsilon \{J ,Q\}~=~-\epsilon, \cr \delta t~=~&0.\end{align} \tag{F} $$