Let $(M,\omega)$ be a compact Kähler manifold and suppose there are holomorpic vector fields vanishing at a point $p$. As a consequence we have a group $G\_{p}$ of biholomorpisms fixing $p$. Let $T\_{p}$ a maximal torus of $G\_{p}$. Suppose $\omega$ is invariant for the action of $T\_{p}$. Can we find a neighborhood $\mathfrak{U}$ of $p$ and a set of holomorphic coordinates $z$ centered at $p$ such that the action of $T\_{p}$ is linear (unitary) **and** coordinates $z$ are Kähler normal coordinates i.e.
\begin{equation}
\omega=i\partial\overline{\partial}\left(\frac{|z|^{2}}{2}+\mathcal{O}(|z|^{4})\right)
\end{equation} 


Separately these two requests can be satisfied, but I can't prove that can hold simultaneously. Is there a reference for this? Or can someone give me a hint? 



Special Kähler normal coordinates around a point
I don't know a reference, but this desired normal form is, indeed, attainable. Here is the argument:


Assume given a Kähler form $\omega$ defined on a neighborhood of $0\in\mathbb{C}^n$ and that there is a torus (i.e., a connected compact abelian Lie group) $\mathbb{T}$ acting effectively and holomorphically on $\mathbb{C}^n$ as $\omega$-isometries and fixing $0\in\mathbb{C}^n$.


As the OP points out, it is known that there exist holomorphic coordinates $z = (z^r)$ centered on $0$ in which the action of $\mathbb{T}$ is *linear*, i.e., that there exist $n$ characters (i.e., homomorphisms) $\chi\_r:\mathbb{T}\to S^1$ such that 
$$
a^\*(z^r) = \chi\_r(a)\,z^r\qquad \text{(no sum)}
$$
for all $a\in \mathbb{T}$.


Now, write $\omega = \mathrm{i}\partial\bar\partial\phi$, where $\phi$ is a Kähler potential for $\omega$ in a neighborhood of $0$. Since $\omega$ satisfies $a^\*\omega = \omega$ for all $a\in\mathbb{T}$, by averaging $\phi$ with respect to $\mathbb{T}$, one can assume that $\phi$ is also $\mathbb{T}$-invariant. After discarding holomorphic and anti-holomorphic terms (which are $\mathbb{T}$-invariant and don't show up in $\omega$), it can be assumed that $\phi$ has a Taylor expansion up to terms vanishing to order $4$ of the form
$$
\phi \equiv\_4 h\_{p\bar q}\,z^p\overline{z^q} + a\_{pq}^r\,z^pz^q\overline{z^r}
 + \overline{a\_{pq}^r}\,\overline{z^p}\overline{z^q}z^r
$$
where $\overline{h\_{p\bar q}} = h\_{q\bar p}$ is positive definite, and $a\_{pq}^r = a\_{qp}^r$. 


Now, because $\phi$ is $\mathbb{T}$-invariant, it follows that $h\_{p\bar q} = 0$ unless $\chi\_p = \chi\_q$. Thus, by collecting the $z^j$ that share the same character and making a 'block-form' linear transformation that preserves these subsets, one can arrange that the $z^p$ have been chosen so that $h\_{p\bar q} = \delta\_{pq}$, i.e., that the quadratic term in $\phi$ is simply $\tfrac12\bigl(|z^1|^2+\cdots+|z^n|^2\bigr)$.


Now, again because of the $\mathbb{T}$-invariance of $\phi$, one sees that $a\_{pq}^r = 0$ unless $\chi\_r = \chi\_p\chi\_q$. Now, consider the change of variables of the form
$$
w^r = z^r + 2a\_{pq}^r\,z^pz^q.
$$
Because $a\_{pq}^r$ vanishes unless $\chi\_r = \chi\_p\chi\_q$, it follows that $a^\*(w^r) = \chi\_r(a)\,w^r$ for all $r$, so the $w$-coordinates also linearize the action of $\mathbb{T}$. Moreover, one clearly has
$$
\phi \equiv\_4 \tfrac12\bigl(|w^1|^2+\cdots+|w^n|^2\bigr).
$$


