Let $X$ be a vector field on the manifold $M$ and let $Y$ be a vector field on $N$. Denote with $e^{tX}$, $e^{tY}$ the flows they generate, that is $$ u(t)=e^{tX}p\quad \text{is the unique solution to }\quad \begin{cases} \dot{u}(t)=X_{u(t)} \\ u(0)=p\end{cases}$$ and similarly for $e^{tY}$. Assume that both $X$ and $Y$ are complete, meaning that $e^{tX}$ and $e^{tY}$ are defined for all $t\in \mathbb{R}$, regardless of the initial condition.
Let $F\colon M \to N$ be smooth. By differentiating at $t=0$, one sees that the property $$\tag{1} F(e^{tX}p)=e^{tY}F(p),\quad \forall p\in M,\ \forall t \in \mathbb{R}$$ implies $$\tag{2} F_{\star{}p}X_p=Y_{F(p)}, \quad \forall p \in M,$$ the term $F_\star$ denoting the push-forward.
Question. Does $(2)$ imply $(1)$?
In the special case $M=\mathbb{R}^n, N=\mathbb{R}$ and $$ e^{tX}p=e^t p,\quad e^{tY}\lambda=e^{k t}\lambda, $$ the equivalence between (1) and (2) holds, and it is called Euler's homogeneous function theorem.