Let $\eta$ be a closed $(1,1)$-form. Then $\partial\eta=0$ and $\overline{\partial}\eta=0$. Recall the Kähler identity \begin{equation*} [\partial,\Lambda] = -i\overline{\partial}^\ast . \end{equation*} Suppose that $\Lambda\eta$ is closed (this follows if it is constant). Then \begin{equation*} 0 = \partial\Lambda\eta = [\partial,\Lambda]\eta + \Lambda\partial\eta = i\overline{\partial}^\ast\eta . \end{equation*} Thus $\overline{\partial}^\ast\eta=0$. Therefore $\eta$ is harmonic.

You already gave the argument showing that if $\eta$ is a harmonic $(1,1)$-form, then $\Lambda\eta$ is a harmonic function. If your Kähler manifold is compact, this implies that $\Lambda\eta$ is locally constant. If $X$ is also connected, then $\Lambda\eta$ is constant.

(Note that similar argument shows that if $\eta$ is a closed $(p,q)$-form, then it is harmonic if and only if $\Lambda\eta$ is harmonic.)

Let $\eta$ be a closed $(1,1)$-form. Then $\partial\eta=0$ and $\overline{\partial}\eta=0$. Recall the Kähler identity \begin{equation*} [\partial,\Lambda] = -i\overline{\partial}^\ast . \end{equation*} Suppose that $\Lambda\eta$ is closed (this follows if it is constant). Then \begin{equation*} 0 = \partial\Lambda\eta = [\partial,\Lambda]\eta + \Lambda\partial\eta = i\overline{\partial}^\ast\eta . \end{equation*} Thus $\overline{\partial}^\ast\eta=0$. Therefore $\eta$ is harmonic.

You already gave the argument showing that if $\eta$ is a harmonic $(1,1)$-form, then $\Lambda\eta$ is a harmonic function. If your Kähler manifold is compact, this implies that $\Lambda\eta$ is locally constant. If $X$ is also connected, then $\Lambda\eta$ is constant.

Without assuming that the Kähler manifold is compact and connected, one cannot say that if $\eta$ is harmonic, then $\Lambda\eta$ is constant. For example, if $(X,\omega)$ is a one-dimensional Kähler manifold and $\eta$ is a $(1,1)$-form, then $\eta = f\omega$ for some function $f$. Moreover, $f = \Lambda\eta$ and $\Delta\eta = (\Delta f)\omega$. Hence $\eta$ is harmonic if and only if $\Lambda\eta$ is harmonic. But there are nonconstant harmonic functions when $X$ is noncompact (e.g. $f=z$) and when $X$ is compact but not connected (e.g. $f=1$ on one component and zero on the others).

Let $\eta$ be a closed $(1,1)$-form. Then $\partial\eta=0$ and $\overline{\partial}\eta=0$. Recall the Kähler identity \begin{equation*} [\partial,\Lambda] = -i\overline{\partial}^\ast . \end{equation*} Suppose that $\Lambda\eta$ is constant. Then \begin{equation*} 0 = \partial\Lambda\eta = [\partial,\Lambda]\eta + \Lambda\partial\eta = i\overline{\partial}^\ast\eta . \end{equation*} Thus $\overline{\partial}^\ast\eta=0$. Therefore $\eta$ is harmonic. (Note that a similar argument works if $\eta$ is a closed $(p,q)$-form.)

You already gave the converse.

(By the way, Siu’s argument is fine, but it is important that the Kähler form $\omega$ is parallel in his statement that $\eta \wedge \omega^{n-1}$ is harmonic.)

Let $\eta$ be a closed $(1,1)$-form. Then $\partial\eta=0$ and $\overline{\partial}\eta=0$. Recall the Kähler identity \begin{equation*} [\partial,\Lambda] = -i\overline{\partial}^\ast . \end{equation*} Suppose that $\Lambda\eta$ is closed (this follows if it is constant). Then \begin{equation*} 0 = \partial\Lambda\eta = [\partial,\Lambda]\eta + \Lambda\partial\eta = i\overline{\partial}^\ast\eta . \end{equation*} Thus $\overline{\partial}^\ast\eta=0$. Therefore $\eta$ is harmonic.

You already gave the argument showing that if $\eta$ is a harmonic $(1,1)$-form, then $\Lambda\eta$ is a harmonic function. If your Kähler manifold is compact, this implies that $\Lambda\eta$ is locally constant. If $X$ is also connected, then $\Lambda\eta$ is constant.

(Note that similar argument shows that if $\eta$ is a closed $(p,q)$-form, then it is harmonic if and only if $\Lambda\eta$ is harmonic.)