Consider the following: We are given a symplectic manifold $M$. Now, we define a Hamilton function $H : M \rightarrow \mathbb{R}.$ Additionally, we want that $H^{-1}(x)=:M_x$ is a submanifold. We can for instance assume that $x$ is a regular value.
Now, the Lie derivative $L_{X_{H}}H = \{H,H\} = 0,$ where $X_H$ is the Hamilton vector field to $H.$
Does this imply that the Hamiltonian flow $\phi^t: M_x \rightarrow M$ is invariant under $M_x$, so do we have $\phi^t: M_x \rightarrow M_x$?
Here $\phi^t$ is the flow to the ODE $x'(t) = X_H(x(t)).$ Intuitively, the question appears trivial, as we are moving on a constant energy manifold, so this should hold.
But how can I see that the flow maps actually maps to $M_x$?
You just need to show that $X_H$ is tangent to $M_x$ (at points in $M_x$). Then its integral curves that intersect $M_x$ lie in $M_x$, as desired.
Because $H$ is constant on $M_x$, $dH$ is zero on $M_x$ (that is, if $Y$ is a tangent vector to $M_x$ at $p \in M_x$, then $dH(Y) = 0$). So because $X_H$ is defined as the unique vector field such that $\omega(X_H, Y) = dH(Y)$, we see that $\omega(X_H, Y) = 0$ for all $Y \in TM_x$. Because $TM_x$ has codimension one in $TM$, it's coisotropic, and thus $X_H \in (TM_x)^\perp \implies X_H \in TM_x$, as desired.
That was just linear algebra. If you want to throw the Lie derivative in there, as in your question, you know that because $L_{X_H} H = 0$, $H$ is invariant under the flow of $X_H$, and hence the flow of $X_H$ preserves the level sets of $H$ as desired.