Q:

symplectic topology of (perturbed) KAM tori

Consider a real analytic $H_0:\mathbb{R}^n\to \mathbb{R}$ whose Hessian is everywhere non-degenerate as well as a real analytic $F:\mathbb{T}^n\times \mathbb{R}^n\to \mathbb{R}$. KAM theory studies what happens to (Lagrangian) tori which are invariant under the Hamiltonian flow $\phi_{H_{\epsilon}}$, associated to 
$$ H_{\epsilon}(q,p)=H_0(p)+\epsilon F(q,p) \quad \forall \ (q,p)\in  \mathbb{T}^n\times \mathbb{R}^n,$$
when $\epsilon$ varies. If the rotation vector of the restriction of the Hamiltonian flow $\phi_{H_{0}}$ to the torus $T(p_0):=\mathbb{T}^n\times \{p_0\}$ is Diophantine, then the KAM theorem (Kolmogorov-Arnold-Moser) guarantees, for all small enough $\epsilon>0$, the existence of an invariant Lagrangian torus $L\approx \mathbb{T}^n \subset \mathbb{T}^n\times \mathbb{R}^n$ which is "close" to $T(p_0)$ and invariant under $\phi_{H_{\epsilon}}$. 
Is $L$ also guaranteed to be Hamiltonian isotopic to $T(p_0)$?
Any explanations/references will be much appreciated.

EDIT: It seems to me that there are two different ways of "parametrizing" invariant sets (at least in Aubry-Mather theory). I agree that, in Richard's example below, it is clear that the tori which have the same rotation rotation vector will not be Hamiltonian isotopic. However, in the Aubry-Mather picture one often "parametrizes" invariant sets by cohomology classes rather than homology classes. Let's denote by $\mathcal{M}_c(\epsilon)$ the $H_{\epsilon}$-Mather set associated to $c\in H^1(T^*\mathbb{T}^n)\cong \mathbb{R}^n$. I.e. $\mathcal{M}_c(\epsilon)$ is the union of supports of invariant probability measures which minimize 
$$ A_c(\mu)=\int L_{\epsilon}d\mu -\langle c,\rho(\mu) \rangle, $$  
where $\rho(\mu)$ denotes the rotation vector of $\mu$ and $L_{\epsilon}$ is the Lagrangian associated to $H_{\epsilon}$ (assuming $H_{\epsilon}$ is "Tonelli").
Then $\mathcal{M}_{p_0}(0)=T(p_0)$. If $\mathcal{M}_{p_0}(\epsilon)$ is diffeomorphic to a torus, is it then true that $\mathcal{M}_{p_0}(\epsilon)$ is Hamiltonian isotopic to $T(p_0)$?   

A:

No. (Answer courtesy of Jacques Féjoz, via email correspondence.) The invariant torus might be translated in the actions,in which case the perturbed torus is not Hamiltonian isotopic to the original.  Consider the trivial, integrable perturbation $H_\epsilon(q,p) = H_0(p+ \epsilon \Delta p_0)$, where $\Delta p_0$ is a fixed vector in action space. Then  the torus  corresponding to $T(p_0)$ -i.e with the same frequency vector, is simply the translate of $T(p_0)$ by $\epsilon \Delta p_0$, and is thus symplectically but not Hamiltonian isotopic to the unperturbed one. This is the general case.  See theorem 20 of https://www.ceremade.dauphine.fr/~fejoz/Articles/Fejoz_2016_introduction-KAM.pdf
where it is proved that the conjugacy is the composition of   Hamiltonian isotopy
by such a translation. 

