The question hopefully says it all!
We have a Hermitian matrix $A=A^* \in \mathbb{C}^n$ and a quadratic form: $f(x)=x^*Ax,~x\in \mathbb{C}^n$
We want to find the solution of $f(x) = x^*Ax = 0$
When the matrix is positive semi definite (p.s.d.), the solution seems to be null-space of the matrix of $A$. This I found by diagonalising $A$. But suddenly I became helpless when $A$ has negative Eigen values too!
Please note that when the matrix is not p.s.d., the solution space contains the null space. (I.e. null space is always a solution)
Consider $N\times N$ matrices and $N\times 1$ vectors. Let $A=U\Lambda U^H$ be its eigen-decomposition. Then \begin{align} x^HAx &=y^H\Lambda y && \{\text{where I define }y=Ux,~\forall x \} \\ &=\sum_{i=1}^{N}|y_i|^2\lambda_i \\ &= \theta^T\lambda \end{align} where $\lambda$ is the vector with all eigenvalues and $\theta$ is any vector whose entries are non-negative. Thus, for any $\theta\in\left(\mathcal{N}(\lambda)\cap\mathbb{R}_{+}^N\right)$ where $\mathcal{N}(\lambda)$ is the null-space of $\lambda$ vector (set of all vectors orthogonal to $\lambda$) and the non-negative $N$-dimensional orthant (non-negative quadrant where all entries are non-negative), we have $\theta^T\lambda=0$. Now consider the set $\mathcal{D}$ of all diagonal matrices such that the diagonal entries lie on the unit circle in the complex plane. Then your solution set is \begin{align} \mathcal{S}_x\,=\,\{U^HD\sqrt{\theta} \,\mid\,\forall D\in\mathcal{D}~,~\forall \theta\in\left(\mathcal{N}(\lambda)\cap\mathbb{R}_{+}^N\right) \} \end{align} where $\sqrt{\theta}$ is the entry-wise square root of $\theta$. $\mathcal{D}$ is needed because it doesn't matter what the phase of each entry of $y$ is. Note that $y_{i}=D_{ii}\sqrt{\theta_i}$ and $|y_i|^2=|D_{ii}|^2\theta_i=\theta_i$. The difficult part is that $\mathcal{S}_x$ is not a linear subspace and you won't have that nice properties. It is a highly non-linear transformation and I am not sure what intuitive sense you can derive out of it.