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Let $\phi$ be a local flow of a vector field $X$ on a Riemannian manifold $(M,g)$. We see that $L_Xg=fg$ if a flow consists of conformal maps because of the formula:

$$(L_Xg)_x=\frac{d}{dt}((\phi_t)^*g)_x|_{t=0}.$$ Unfortunatelly I have a problem with showing inverse i.e. that $L_Xg=fg$ implies that flow consists of conformal maps.

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There are two main ingredients to the proof.

Lemma 1. Suppose $V$ is a vector space and $g(t)$ is a tensor on $V$ depending smoothly on a parameter $t\in J$ for some interval $J\subseteq\mathbb R$ containing $0$. If $$\frac{d}{dt} g(t) = u(t)g(t)$$ for some smooth function $u\colon J\to \mathbb R$, then there is a smooth function $F\colon J\to \mathbb R$ such that $g(t) = e^{F(t)} g(0)$.

Proof. Let $F(t) = \int_0^t u(s)\,ds$, so that $F(0)=0$ and $F'(t)=u(t)$. Direct computation shows that the derivative of $e^{-F(t)}g(t)$ is identically zero, so it is equal to its value at $t=0$, which is $g(0)$. $\square$

Lemma 2. Let $M$ be a smooth manifold, $X$ be a smooth vector field on $M$ with flow $\phi$, and $g$ be a smooth tensor field on $M$. For each $t_0\in\mathbb R$ such that $\phi_{t_0}$ is defined, $$ \left.\frac{d}{dt}\right|_{t=t_0} \phi_t^* g = \phi_{t_0}^* \left( \mathcal L_X g\right). $$

Proof. This is Proposition 12.36 in my Introduction to Smooth Manifolds (2nd ed.). $\square$

Now to answer your question:

Theorem. Let $X$ be a vector field on a Riemannian manifold $(M,g)$, and let $\phi$ be its flow. If $\mathcal L_Xg=fg$ for some smooth function $f\colon M\to\mathbb R$, then $\phi_t$ is a conformal map for each $t$.

Proof. Let $p$ be an arbitrary point in $M$, and define a time-dependent tensor $g(t)$ on $T_pM$ by $g(t) = (\phi_t^*g)|_p$. Then Lemma 2 shows that for each time $t_0$, $$ \left.\frac{d}{dt}\right|_{t=t_0} g(t) = \big(\phi_{t_0}^* \left(f\, g\right)\big)\Big|_p = (f\circ\phi_{t_0}(p)) \phi_{t_0}^* g\Big|_p = (f\circ\phi_{t_0}(p))g(t_0). $$ The result follows by applying Lemma 1 with $u(t)=f\circ\phi_{t}(p)$. $\square$

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