First, the scaling you mention amounts to failure to clean up your variables,
by defining
$$
\tilde p\equiv \frac{p}{m\omega},
$$
likewise for $\tilde P$, and
$$
\tilde H \equiv \frac{H}{m\omega^2}= \frac{1}{2} (\tilde p^2+ q^2).
$$
It is then apparent that the transformation
$$
Q=q \cos \theta -\tilde p \sin\theta\\
\tilde P= \tilde p \cos\theta + q \sin\theta, \tag{0}
$$
the rotation you observed, is a continuous symmetry of your $\tilde H$, a rotational scalar.
It is also manifestly a canonical transformation without even consideration of generating functions, wont to confuse you, since it preserves Poisson Brackets, essentially by inspection,
$$
\{ Q,\tilde P \}=1, $$
(of course, $\{ Q,Q \}=0, \quad \{ \tilde P,\tilde P \}=0$.)
But, in your particular case, the generator which produces your canonical transformation by Lie Poisson commutation,
$$
Q= e^{\{ G, \bullet \}} ~~ q = q +\{ G, q \} + \frac{1}{2} \{ G , \{ G, q \} \} +\frac{1}{3!}\{G,\{ G, \{ G, q\}\}\}+ ... \\
\tilde P = e^{\{ G, \bullet \}} ~~ \tilde p = \tilde p +\{ G, \tilde p \} + \frac{1}{2} \{ G , \{ G, \tilde p \} \} +... \tag{1}
$$
happens to be
$$
G= \theta \tilde H,
$$
so, of course, it Poisson-commutes with itself, $\{ G, \tilde H \}=0$. (Do check that, since $\{G,q\}=-\theta \tilde p$, etc, the above series sum to the exact finite rotation (0).) Consequently, the transformation it generates leaves the Hamiltonian invariant. Your criterion is Poisson-commutation with the Hamiltonian.
- But this is exactly how this Hamiltonian, through Lie iteration of Hamilton's equations, acts to generate motion, here a mere phase-space rotation. (Coincidentally, here, you are replicating the general fact that motion is a canonical transformation, crucially important in QM.)
To sum up, in this language, given a generator $G(q,\tilde p)$ Poisson-commuting with the Hamiltonian, it will be a symmetry thereof, so the (provably canonical, below) transformations it generates will leave the Hamiltonian invariant.
I am not sure I understand your second point, since I cannot fathom your "no benefits to the problem". Canonical changes of variables allow you to simplify problems to the point where their solutions are virtually self-evident. Have you gotten to the action-angle variables part of your course? The fact that motion is a canonical transformation and preserves the local dynamics of the theory in question is also central.
- Proof of the generic finite transformation (1) being canonical, as per comment. (Geeky)
For generic $G=\theta g(q,p)$, not just the above, is the following quantity =1?
$$
Y(\theta)= \{Q,P \} \equiv \{ e^{\{\theta g , \bullet \}} ~q ~,~ e^{\{ \theta g, \bullet \}} ~p \} \\ \equiv 1+\theta y_1
+\theta^2 y_2+\theta^3 y_3+ ...
$$
Consider, for any $\theta$-independent function $f(q,p)$,
$$
\frac{\partial (e^{\{ \theta g, \bullet \}}~ f)} {\partial \theta}= \{ g ~,~ e^{\{ \theta g, \bullet \}} ~f \},
$$
implying, by the Jacobi identity,
$$
\frac{\partial Y(\theta) } {\partial \theta}=
\{\{ g,Q\},P\}+ \{Q,\{g,P\}\}=\{g,Y(\theta)\}.
$$
So, then, $y_{n+1}=\{ g,y_n\}$, constraining all higher coefficients to zero, since $y_1=\{g,1\}=0$, hence $Y(\theta)=1$, independent of $\theta$.
In the same breath, you may prove the all-orders converse Noether's theorem, $\{g,H\}=0 \Longrightarrow H(Q,P)=H(q,p)$, paradoxically easier to prove in QM (!), but available in Arnold's classic text. Here, I will just all but remind you the lowest-order result is evident by inspection,
$$
H(Q,P)=H(q,p)+ \theta ~ \left (-\frac{\partial H}{\partial q} \frac{\partial g}{\partial p}+\frac{\partial H}{\partial p}\frac{\partial g}{\partial q} \right )+ O(\theta^2)\\
= H(q,p)+ \theta ~ \{ g,H\} + O(\theta^2)=H(q,p)+ O(\theta^2).
$$
The $O(\theta^2)$ vanishes as well, since, by $\{Q,P\}=1$, it does not matter which basis our PBs are in; and hence, likewise and by the above proof, $\partial H(Q,P)/\partial \theta= ... = \{g,H\}=0$, to all orders in $\theta$.
