Let $\varphi:\mathbb R\times \mathbb R^n\to \mathbb R^n$ be a flow, that is, a $C^\infty$ map such that $\varphi(0,p) = p$ for every $p\in\mathbb R^n$ and $\varphi(t+s,p) = \varphi(s,\varphi(t,p))$ for every $p\in \mathbb R^n$ and $t,s\in\mathbb R$.
Lets suppose that for every $t\in\mathbb R$, the map $\varphi_t:\mathbb R^n\to \mathbb R^n$, $\varphi_t = \varphi(t,\cdot)$ is a linear map, so $\varphi_t$ is in particular a diffeomorphism. I want to prove that there exist a $n\times n$ real matrix $A$ such that $\varphi_t = e^{tA}$, so I would be very pleased if someone gives me a hint. Thanks in advance.
Edit: Since $e^{tA}$ needs to appear in some way, and it is the solution of the ordinary differential equation system $y'(t) = A y(t)$, when $y(0) = 1$.
Hint(Rough Sketch): Observe for each fixed $t$ we have \begin{align} \varphi^t(x+y) = \varphi^t(x)+\varphi^t(y) \end{align} which is a Cauchy functional equation where $\varphi^t$ is continuous. Observe, if \begin{align} \varphi^t(nx) = n\varphi^t(x) \end{align} for all $n \in \mathbb{Z}$ which leads us to show that \begin{align} \varphi^t(q x)= q\varphi^t(x) \end{align} where $q \in \mathbb{Q}$. By continuity, it follows $\varphi^t(\alpha x) = \alpha \varphi^t(x)$ for all $\alpha \in \mathbb{R}$. Moreover, consider the standard basis for $\mathbb{R}^n$, then we see that \begin{align} \varphi^t(x) = \varphi^t(\alpha_1 e_1 + \ldots +\alpha_n e_n) = \sum^n_{k=1}\alpha_k \varphi^t(e_k) \end{align} which means \begin{align} \varphi^t(x) = \begin{pmatrix} \varphi^t(e_1), \ldots, \varphi^t(e_n) \end{pmatrix} \begin{pmatrix} \alpha_1\\ \alpha_2\\ \vdots\\ \alpha_n \end{pmatrix} =A(t)x. \end{align} Hence we now have \begin{align} \varphi^t(x) = A(t)x \ \ \Rightarrow \ \ \frac{d}{dt}\varphi(t, x) = A'(t)x = A'(t)A^{-1}(t)\varphi(t, x) = B(t)\varphi^t(x). \end{align} Thus, solving the above differential equation in $t$ yields \begin{align} \varphi(t, x) = e^{\int^t_0 B(\tau)\ d\tau}x. \end{align} Finally, using the property \begin{align} \varphi(t, \varphi(s, x)) = \varphi(t+s, x) \end{align} we see that \begin{align} e^{\int^t_0 B(\tau)\ d\tau}\left[e^{\int^s_0 B(\tau)\ d\tau}x \right] = e^{\int^{t+s}_0 B(\tau)\ d\tau}x. \end{align} Denote \begin{align} f(t) = \int^t_0B(\tau)\ d\tau \end{align} then it follows we have the following functional equation \begin{align} f(t)f(s) = f(t+s) \end{align} which leads to \begin{align} f(t) = e^{tA} \end{align} where $A$ is a constant matrix.