Gaussian fourth-moment formulas? Suppose $X$ is an $m\times n$ random matrix where rows are I.I.D. samples of $n$-dimensional Gaussian, and $A$ is an $n\times n$ matrix. I'm looking for the value of the following quantities:
$$E[X^T X A X^T X]$$
For zero-centered Gaussian with covariance $\Sigma$ and $m=1$, the following holds
$$E[X^T X A X^T X]=\\
\Sigma A \Sigma + \Sigma A^T \Sigma + \Sigma \operatorname{Tr}(\Sigma A)$$
What is the expression for more general case?
Edit
The following formula from the answer has been confirmed to work:
$$\begin{equation}
\frac{1}{m^2}E[X^TXAX^TX] = \\\frac{(m-1)}{m}HAH + \frac{1}{m}
\left( H(A+A^T)H - \mu\mu^T(A+A^T)\mu\mu^T + H\operatorname{Tr}(HA) \right),
\end{equation}$$
where $H=\Sigma+\mu\mu^T$
 A: I have never seen an closed form expression for this. Probably because it is quite ugly. I have worked with a similar expression before, and I'd be happy to see if my expression is stands up to yours. So here is one way to compute the matrix expectation.
I will work with $\frac{X^TX}{m}$ rather than $X^TX$, since that is the empirical 2nd order moment matrix. It is a quantity that will converge.
Lemma
First some tricks that need index manipulations for me to see. If $y\sim \mathcal{N}(0,\Sigma)$, $\mu$ a vector and $A$ is a matrix,
$$E[y\mu{}^T A y \mu^T] = \Sigma{}A^T\mu\mu^T$$
$$E[\mu{}y^TA\mu{}y^T] = \mu\mu^T A^T \Sigma $$
Proof:
Follow the indices! I use Einstein summation convention to be able declutter notation.
$$E[y\mu{}^T A y \mu]_{ij} = E[y_i\mu_kA_{kl}y_l\mu_j] = E[y_iy_l]A^T_{lk} \mu_k \mu_j= \Sigma_{il}A^T_{lk}\mu_k\mu_j = [\Sigma{}A^T\mu\mu^T]_{ij}$$
The second equation is similarly proven.
QED
Next, we will look at the zero-mean case as a warm-up exercise.
Lemma
Assume $m$ independent gaussian random vectors $x^{(i)} \overset{iid}{\sim} \mathcal{N}(0,\Sigma)$. Row-stack them in the matrix $X$. Then
\begin{equation}
    \begin{split}
E\left[\frac{X^TX}{m}A\frac{X^TX}{m}\right] &=    \frac{m-1}{m} \Sigma{}A\Sigma{} \\
& + \frac{1}{m}\left( \Sigma{}(A+A^T)\Sigma{} + \operatorname{Tr}(\Sigma{}A)\Sigma{} \right)
    \end{split}
\end{equation}
Proof:
$$
    \frac{1}{m^2}E[X^TXAX^TX] = \sum_{i,j}E[x^{(i)}{x^{(i)}}^TAx^{(j)}{x^{(j)}}^T]
$$
There are $m(m-1)$ terms where $i\neq{}j$, and $m$ terms where they agree.
Since $x^{(i)}$ and $x^{(j)}$ are iid, we can let $x$ denote either of them, and state
$$
    \frac{1}{m^2}E[X^TXAX^TX] = \frac{1}{m^2}( m(m-1) E[xx^T]AE[xx^T] + m E[xx^TAxx^T])
$$
The first term is just the variances. The second term needs isserlis theorem. It gives (again, einstein summation convention)
\begin{align}
E[xx^TAxx^T]_{ij} &= E[x_ix_kA_{kl}x_lx_j] \\&= A_{kl}(E[x_ix_k]E[x_lx_j]+E[x_ix_l]E[x_kx_j]+E[x_ix_j]E[x_lx_k] \\&= \left\{ \Sigma(A+A^T)\Sigma + \Sigma \operatorname{Tr}(A\Sigma)\right\}_{ij}
\end{align}
Armed with this, we can simply state
$$
    \frac{1}{m^2}E[X^TXAX^TX] = \frac{m-1}{m^2} \Sigma{}A\Sigma{} + \frac{1}{m}(\Sigma(A+A^T)\Sigma + \Sigma \operatorname{Tr}(A\Sigma))
$$
QED
Now, we look at the general case.
Lemma:
Assume $m$ independent gaussian random vectors $x^{(i)} \overset{iid}{\sim} \mathcal{N}(\mu,\Sigma)$. Row-stack them in the matrix $X$. Let $H=\Sigma+\mu\mu^T$.
Then
\begin{equation}
E\left[\frac{X^TX}{m}A\frac{X^TX}{m}\right] = \frac{(m-1)}{m}HAH + \frac{1}{m}
\left( H(A+A^T)H - \mu\mu^T(A+A^T)\mu\mu^T + H\operatorname{Tr}(HA)) \right),
\end{equation}
Proof:
Let $x \sim \mathcal{N}(\mu,\Sigma)$. Define $y=x-\mu$.
As before,
\begin{equation}
\frac{1}{m^2}E[X^TXAX^TX] = \frac{1}{m^2}( m(m-1) E[xx^T]AE[xx^T] + m E[xx^TAxx^T]) \label{eq:one}
\end{equation}
The first term is simply
\begin{equation}
\label{eq:two}
E[xx^T]AE[xx^T] = (\Sigma+\mu\mu^T)A(\Sigma+\mu\mu^T).
\end{equation}
The second term needs more work. We expand it in $y+\mu$, and notice that all terms with 1 or 3 factors of $y$ must have expectation 0 due to either Isserlis theorem, or to the zero mean. There are in total 16 terms, but 8 gets zeroes out in that way.
\begin{equation}
\begin{split}
E[xx^TAxx^T] &= E[(y+\mu) (y+\mu)^TA(y+\mu)(y+\mu)^T] \\
& = E[yy^TAyy^T] + E[\mu\mu^TA\mu\mu^T]\\
& + E[yy^TA\mu\mu^T] + E[\mu\mu^TAyy^T]\\
& + E[y\mu^TAy\mu^T] + E[\mu{}y^TA\mu{}y^T]\\
& + E[y\mu^TA\mu{}y^T] + E[\mu{}y^TAy\mu{}^T]
\end{split}
\end{equation}
All these are simple expectations. The first one due to Isserlis' theorem. The second is the expectation of a constant. Term three and four are just covariances. Term five and six are computed by the initial Lemma that I needed some index manipulations for. Term seven is due to the inner three factors are scalar so the $y$'s can be combined. Term eight uses $y^TAy = \operatorname{Tr}(yy^TA)$ and that the trace is linear and thus commutes with expectations. So, we find
\begin{equation}
\begin{split}
E[xx^TAxx^T] &= (\Sigma(A+A^T)\Sigma + \Sigma \operatorname{Tr}(A\Sigma)) + \mu\mu^TA\mu\mu^T\\
& + \Sigma{}A\mu\mu^T + \mu\mu^TA\Sigma{}\\
& + \Sigma{}A^T\mu\mu^T + \mu\mu^TA^T\Sigma \\
& + \Sigma\mu^TA\mu  + \mu\mu^T\operatorname{Tr}(\Sigma{}A)
\end{split}
\end{equation}
Some regrouping of terms yields
\begin{equation}
\begin{split}
E[xx^TAxx^T] &= (\Sigma+\mu\mu^T)(A+A^T)(\Sigma+\mu\mu^T) - \mu\mu^T(A+A^T)\mu\mu^T\\
& + (\Sigma+\mu\mu^T)\operatorname{Tr}((\Sigma+\mu\mu^T)A) 
\end{split}
\end{equation}
Combining the found results, we get
\begin{equation}
E\left[\frac{X^TX}{m}A\frac{X^TX}{m}\right] = \frac{(m-1)}{m}HAH + \frac{1}{m}
\left( H(A+A^T)H - \mu\mu^T(A+A^T)\mu\mu^T + H\operatorname{Tr}(HA)) \right),
\end{equation}
where $H=\Sigma+\mu\mu^T$
QED
