Given a Poisson manifold $(M,\{\cdot,\cdot\})$ and any function $f\in C^\infty(M)$, the corresponding flow of the hamiltonian vector field $X_f$ at some instant $\tau$, this is, $\phi^\tau_{X_f}:M\to M$, is a Poisson map?
And a Poisson map is defined as a function $ \varphi:M_1\to M_2$ such that $$\{h\circ\varphi,g\circ\varphi\}_1= \{h,g\}_2\circ\varphi , \ \forall h,g\in C^{\infty}(M_2).$$
Then, the question here is if $\{h\circ\phi^\tau_{X_f},g\circ\phi^\tau_{X_f}\}= \{h,g\}\circ\phi^\tau_{X_f} $ for any functions $f,g,h\in C^\infty(M)$.
Thanks in advance.
The result is a bit more general. For every vector field which is locally hamiltonian (and hence, hamiltonian), the Lie derivative of the Poisson bivector is $0$. Let $X$ be a locally hamiltonian vector field. Then, for any open $U$ of $M$, there is a function $f\in C^\infty(U)$ such that $X|_U=\{f,\cdot\}|_U$. Let $h,g\in C^\infty(U)$, then, using the usual formula for the Lie derivative and the antisymmetry of the Poisson bracket \begin{gather*}\mathcal{L}_X\pi(h,g)=X(\pi(h,g))-\pi(X(h),g)-\pi(h,X(g))=\\=\{f,\{h,g\}\}-\{\{f,h\},g\}-\{h,\{f,g\}\}=\{f,\{h,g\}\}+\{g,\{f,h\}\}+\{h,\{g,f\}\}. \end{gather*} And $\{f,\{h,g\}\}+\{g,\{f,h\}\}+\{h,\{g,f\}\}=0 $, by the Jacobi identity.
This proves that $\pi=\{\cdot,\cdot\}$ is constant along the flow $\phi_X$ of any locally hamiltonian vector field $X$, then, only the values that the functions take are ''relevant'' when evaluating the bracket, this is, $\{h,g\}\circ \phi_X=\{h\circ\phi_X,g\circ\phi_X\}$.