Given some $n \times 1$ vector $\mathbf{x}$, with the corresponding $n \times n$ covariance matrix $\Sigma_x$, is there a straightforward way to find the $n^2 \times n^2$ covariance matrix of:
$\mathbf{x} \mathbf{x}^\mathrm{T}$
in terms of the original vector $\mathbf{x}$ and covariance matrix $\Sigma_x$? The elements of $\mathbf{x}$ cannot be assumed to be uncorrelated.
Note that the vectorization of the matrix $V = \mathbf x \mathbf x^T$ is the $n^2 \times 1$ vector $\mathbf v = \mathbf x \otimes \mathbf x$, where $\otimes$ denotes a Kronecker product. With that, we have $$ \mu_\mathbf v = \operatorname{vec}(\mu_{\mathbf x \mathbf x^T}) = \operatorname{vec}(\Sigma_{ \mathbf x}), $$ and $$ K_{\mathbf v \mathbf v} = E(\mathbf v \mathbf v ^T) - \mu_\mathbf v \mu_\mathbf v^T\\ = E((\mathbf x \otimes \mathbf x)(\mathbf x \otimes \mathbf x)^T) - \operatorname{vec}(\Sigma_{ \mathbf x})\operatorname{vec}(\Sigma_{ \mathbf x})^T \\ = E((\mathbf x \mathbf x^T) \otimes (\mathbf x \mathbf x^T)) - \operatorname{vec}(\Sigma_{ \mathbf x})\operatorname{vec}(\Sigma_{ \mathbf x})^T. \\ $$
Note that the components of the first expectation will consist of $4$th moments which cannot be expressed in terms of the entries of $\Sigma_{\mathbf x}$. However, these can perhaps be expressed in terms of the kurtosis and cokurtosis.