Apparently, when we denote by $X(s,t,x)$ the (unique) solution of the initial value problem
\begin{equation} \frac{dX}{dt} = f(X),\ \ \ X(s)=x,\ \ \ s,t \in [0,T],\ x \in \mathbb{R}^n, \end{equation}
then
\begin{equation} (1)\ \ \ \ \ \ \ \ \ \ X(t_3,t_1,x) = X(t_3,t_2,X(t_2,t_1,x)),\ \ \ \forall t_1,t_2,t_3 \in [0,T]. \end{equation}
But unfortunately, I don't see why.
$\textbf{My ideas}$: Probably this has a really simple explanation, for example in the autonomous case a corresponding property follow immediately by showing that both functions satisfy the same initial value problem and thus coincide. Here, we immediately see by choosing $t_2,t_3$ arbitrarily and $t_1 := t_2$ that
\begin{equation} X(t_3,t_1,x)= X(t_3,t_2,x) = X(t_3,t_2,X(t_2,t_2,x)) = X(t_3,t_2,X(t_2,t_1,x)), \end{equation}
so the statement follows, when the function on the right-hand side of (1) solves the same ODE (in $t_1$) as the one on the left-hand side. Unfortunately, all I see
\begin{equation} \frac{dX(t_3,t_2,X(t_2,t_1,x))}{dt_1} = \nabla X (t_3,t_2,X(t_2,t_1,x)) \frac{dX(t_2,t_1,x)}{dt_1} = \nabla X (t_3,t_2,X(t_2,t_1,x)) f(X(t_2,t_1,x)), \end{equation}
and I don't get why this is equal to $f(X(t_3,t_2,X(t_2,t_1,x)))$. Can someone tell me what I'm doing wrong or give me some reference for a proof of this? Thank you