Let $(M,\omega)$ be a $2n$-dimensional symplectic manifold. The claim of Darboux's theorem is that for any point $p\in M$, there exists a local coordinate $(U;q^1,...,q^n,p_1,...,p_n)$, and on $U$, the symplectic form $\omega$ can be written in the form $\omega|_U = dq^i \wedge dp_i$.
I have learned that Darboux's theorem can be paraphrased as follows:
For any points $p\in M, p'\in M'$ in two symplectic manifolds $(M,\omega),(M',\omega')$ of equal dimension, there exists open sets $U, U'$ containing $p,p'$ respectively, and a homomorphic map $\phi : U \rightarrow U'$, satisfying $\phi(p) = p'$ and the pull back $\phi^*\omega' = \omega$.
However, I have tried to think of the proof that the two claims are equivalent, but I can't come up with it, nor have I found books with the proof. Could you please give me the proof or let me know the book where the proof is written?
This answer expands on @J.V.Gaiter's comment. Let $\varphi\colon U \subset \Bbb R^{2n} \to \varphi(U) \subset M$ and $\varphi' \colon U' \subset \Bbb R^{2n} \to \varphi'(U')\subset M'$ be Darboux charts centred at $0$ for $(M,\omega)$ and $(M',\omega')$ respectively. This means that $$ (\varphi^*\omega)|_U = {\omega_0}|_U $$ and $$ ((\varphi')^*\omega')|_{U'} = {\omega_0}|_{U'} $$ where $\omega_0 = \sum_{j=1}^n dx_j\wedge dy_j$ on $\Bbb R^{2n}$ with coordinates $(x_1,y_1,\ldots,x_n,y_n)$. This is precisely the statement of Darboux Theorem applied to $(M,\omega)$ and $(M',\omega')$. Then $(\varphi^*\omega)|_{U\cap U'} = ((\varphi')^*\omega')|_{U\cap U'}$. Notice that $$ \psi =\varphi'\circ \varphi^{-1} \colon \varphi(U\cap U')\subset M \to \varphi'(U\cap U') \subset M' $$ is a diffeomorphism between open subsets of $M$ and $M'$, such that $ \psi^*( \omega'|_{\varphi'(U\cap U')}) = \omega|_{\varphi(U\cap U')}. $
In other words, all $2n$-dimensional symplectic manifolds are locally symplectomorphic. This in particular rules out the existence of local invariants for symplectic geometry.