Let $\Phi: G \times M \rightarrow M$ be a group action on a symplectic manifold $M$ and $G$ be a Lie group.
Furthermore, $x$ is a solution of the Hamilton equation $\dot{x}(t) = X\_H(x(t))$ and for a any $t$ there is a $g(t) \in G\_x:=\{g \in G; Ad^\*\_h(x)=x\}$ (here $x \in \mathfrak{g}^\*$) such that $x(t) = \Phi(g(t),d(t)).$ 


Now this $d:I \rightarrow M$ is another path in $M$ that is not necessarily Hamiltonian, but can be transformed into one under the group action, where the group elements can be taken from $G\_x.$


In this situation I found that $d$ already satisfies the Hamilton equation 


$$\dot{d}(t) + Z\_{\zeta(t)}(d(t))= X\_H(d(t)),$$ 


where $Z\_{\zeta(t)}(p) := \frac{d}{dt}|\_{t=0} \Phi(e^{t \zeta(t)},p)$ and $\zeta(t):=dL\_{g(t)^{-1}}\dot{g}(t).$


So actually $d$ satisfies a Hamiltonian equation with an additional term. Does anybody know how this result can be derived?


It should not be too hard, as we already have a strong relationship between $x$ and $d$, but I don't really see how we can get the equation modified Hamilton's equation shown above.



Non Hamiltonian vector field
I'm afraid I'm not entirely clear on what you are asking - you say you "found" a result, but then ask how it can be derived. Here's a comment that might be helpful.



> 
> In this situation I found that $d$ already satisfies the Hamiltonian equation
>  $$\dot{d}(t)+Z\_{\zeta(t)}(d(t)) = X\_H(d(t)),$$
> 
> 
> 


From your hypothesis, you should get 
$$
\dot{d}(t)+Z\_{\zeta(t)}(d(t)) = d\Phi\_{g(t)^{-1}}(X\_H(x(t))).
$$
Assuming $\Phi$ is a symplectic action, we have that $d\Phi\_{g^{-1}}\cdot X\_{H}(x) = X\_{H\circ \Phi\_g}(d)$. If $H$ is $G$-invariant, this gives the result quoted above.


Beyond that, maybe you could clarify your question (or give some context for it)?


