Well, if the metric $g$ is Kähler and real-analytic, then having the exponential map from even one point $p$ be holomorphic makes it flat. (Probably the assumption of real-analytic is not necessary, but I haven't thought about that, and the following argument that uses it is very easy.)
It's enough to show that a Kähler metric on a neighborhood of $0\in\mathbb{C}^n$ with the property that the linear holomorphic coordinates are Gauss normal coordinates is necessarily flat. To see this, suppose that $z^i$ are linear coordinates on $\mathbb{C}^n$ such that the given metric $g$ agrees with the standard one at $0$. Then let $f$ be a Kähler potential for $g$, so that $$ g=\frac{\partial^2f}{\partial z^i\ \partial\overline{z}^j}\ dz^i d\overline{z}^j. $$ The condition that the $z^i$ be Gauss normal coordinates centered on $0$ is just that $$ z^i\ \overline{z}^j\ \frac{\partial^2f}{\partial z^i\ \partial\overline{z}^j} = |z^1|^2 + \cdots + |z^n|^2. $$ However, now assuming that $f$ is real-analytic, expand it in a power series in $z^i$ and $\overline{z}^i$. Then this last identity clearly shows that all of the terms in $f$ must be zero except the pure $z$-terms, the pure $\overline{z}$-terms, and the terms of bidegree $(1,1)$, which must simply be $|z^1|^2 + \cdots + |z^n|^2$. Thus, $g$ is the standard metric.