Derivative of a vector field is anti-symmetric $\iff$ $dF_t(p)$ is an isometry

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Let $F$ a complete vector field of class $C^1$ in $\Bbb{R}^n$. I need to show that $dF(p)$ is an anti-symmetric matrix $\iff dF_t(p)$ is an isometry, that is

$$\left<dF_t(p)u,dF_t(p)v\right>=\left<u,v\right>\forall\, t\in\Bbb{R}, p,u,v\in\Bbb{R}^n. $$

Here, $F_t(p)$ is the trajectory of $F$ passing through $p$.

I found some answers, like this one, where $F$ is a linear field. In that case, $F_t(p)$ is just an exponential matrix, and all goes fine. But I couldn't addapt the proof for the non linear case.

If someone could give me some hints (not the complete answer) I would appreciate.