Question about calculation the equality of Hamiltonian vector field In McDuff&Salamon's Book, they claimed that since $X_{g^{-1}\xi g}=\psi^*_g X_\xi$,  hence the functions $H_{g^{-1}\xi g}$ and  $\psi^*_g H_\xi$  generate the same Hamiltonian vector field.
Here, $\iota(X_ξ)ω = dH_ξ$, below is my calculation, I wonder what is my fault.

By the definition, $\iota(X_{g^{-1}\xi g})\omega=dH_{g^{-1}\xi g}$.
Assume $\iota(Y)\omega=d\psi^*_g H_\xi$, then
$X_{g^{-1}\xi g}=Y$ $\Leftrightarrow$ $dH_{g^{-1}\xi g}=d\psi^*_g H_\xi$
$\Leftrightarrow$ $\iota(X_{g^{-1}\xi g})\omega=d\psi^*_g H_\xi$
$\Leftrightarrow$ $\iota(\psi^*_g X_\xi)\omega=d\psi^*_g H_\xi$.
Now we verify the last equation:
we have known that $\iota(X_\xi)\omega=dH_\xi$, then
$\psi^*_gdH_\xi(Y)=\psi^*_g\iota(X_\xi)\omega(Y)=\iota(X_\xi)\omega(\psi_{*,g}Y)=\omega(X_\xi,\psi_{*,g}Y)=\psi^*_g\iota(\psi_{*,g^{-1}}X_\xi)\omega(Y)$,
$(\psi^*_g dH_\xi)(Y)=d(\psi^*_g H_\xi)(Y)$,

So I can only get $\psi^*_g\iota(\psi_{*,g^{-1}}X_\xi)(Y)=(d\psi^*_g
 H_\xi)\omega(Y)$ (1), there is an additional $\psi^*_g$.
I wonder either my
calculation is wrong, or, we can earn conclude what we want (equality
of Hamiltonian vector field) from (1).
 A: Your calculation could be somewhat simplified, but the only thing you need is to realize that
$$
(\psi_g^* dH_\xi)(Y) = dH_\xi(\psi_{g,*}Y) = d(H_\xi \circ \psi_g)(Y) = d(\psi_g^* H_\xi)(Y)
$$
and therefore $\psi_g^* dH = d(\psi_g^*H_\xi)$. The key fact is that pullback acts on functions by composition from behind.
EDIT:
Assumption: $X_{g^{-1}\xi g} = \psi_g^* X_\xi$ for all $g\in G$ acting by symplectomorphisms on $M$ and for all $\xi \in Lie(G)$.
Statement: then $H_{g^{-1}\xi g} = \psi_g^* H_\xi$.
Proof: the fundamental vector field $X_{g^{-1}\xi g}$ satisfies $\iota(X_{g^{-1}\xi g})\omega = d(H_{g^{-1}\xi g})$. Then, the assumption is equivalent to
$$
\iota(Y)d(H_{g^{-1}\xi g}) = \iota(Y)\left[\iota(\psi_g^*X)\omega \right]
$$
for all $Y$ vector field. The last is in turn equivalent to
$$
\iota(Y)d(H_{g^{-1}\xi g}) = \omega(\psi_g^*X,Y) \qquad (3)
$$
but by $G$-equivariance of $\omega$ we have that the RHS in the last equation can be rewritten as $\omega(X,\psi_{g^{-1}}^*Y)$ which is $\iota(\psi_{g^{-1}}^*Y)d(H_\xi)$. By the properties of pullbacks and pushforwards, this is
$$
\iota(\psi_{g^{-1}}^* Y)dH_\xi = \iota(Y)d(H_\xi \circ \psi_g)
$$
and thus from (3) we get
$$
\iota(Y)d(H_{g^{-1}\xi g})=\iota(Y)d(H_\xi \circ \psi_g)
$$
for all $Y$, and thus
$$
d(H_{g^{-1}\xi g})=d(H_\xi \circ \psi_g)
$$
Assuming $M$ is a connected manifold, if two functions have the same differential then they are equal up to a constant. If that constant can be chosen to be zero consistently for $\xi \in Lie(G)$, then $H_\cdot$ is called a moment map, which I suppose is introduced in McDuff & Salamon.
$$
H_{g^{-1}\xi g}=H_\xi \circ \psi_g = \psi_g^* H_\xi
$$
Remark: I have been sloppy with pushforwards because I assume $g$ are bijective symplectomorphisms, and everything works fine.
