Let $(M,J)$ be a complex manifold with complex structure $J$. Suppose $X$ is a real vector field over $M$.
$\mathcal{L}_XJ=0$ iff flow of $X$ consists of holomorphic transformation of $M$
The following is a dumb question. The reference is http://moroianu.perso.math.cnrs.fr/tex/kg.pdf
pg 14 Lemma 2.7.
Q: Why this is tautological here? I do not see this is obvious. The flow is made of 1 parameter group along some other real vector field.
This is a general fact about flows and tensors that is essentially a generalization of the fact that a differentiable function is constant if and only its derivative is zero. Whether or not you label it as tautological is a matter of taste, I suppose.
The Lie derivative of a tensor field is typically defined by $$\mathcal L_X T = \frac{d}{dt}\Big|_{t=0} \Phi_t^* T$$where $\Phi_t$ is the diffeomorphism given by flowing along $X$ for time $t$.
Proof. The "if" direction is immediate from the definition above. To get the other direction, we need to "integrate" $\mathcal L_X T = 0.$ Differentiating the flow group law $\Phi_t \Phi_s = \Phi_{s+t}$ with respect to $s$ at $s=0,$ we find that the derivative of $\Phi^*_t T$ at a general time is simply $$\frac{d}{dt} \Phi_t^* T = \Phi_t^* \mathcal L_X T;$$ so $\mathcal L_X T = 0$ in fact implies that $\frac{d}{dt}\Phi_t^* T = 0$ for all times. Thus $\Phi_t^* T$ is constant in $t,$ so it is always equal to $\Phi_0^*T = T.$$\;\square$
Since holomorphic transformations are defined exactly as diffeomorphisms $\phi$ satisfying $\phi^* J = 0,$ this specializes to answer your question.