Let $x_i\in\mathbb{R}^N$ be multivariate normal distributed with mean vector $\mu\in\mathbb{R}^N$ and no correlation: $x_i\sim N(\mu,\sigma^2 I_N)$. Given $T$ iid samples, define the matrix $$X:=\left( \begin{array}{c} x_1' \\ \vdots \\ x_T'\end{array} \right)\in\mathbb{R}^{T \times N}$$ I am interested how the resulting distribution of the elements of the matrix $X'X$ can be characterized. Clearly, the diagonal element $X'X_{i,i}$ contain the sum of independent squared normal variables with mean vector $\mu_i$ and variance $\sigma^2$. Therefore $\sigma^2X'X_{i,i}$ corresponds to a non-central chi squared distribution with $T$ degrees of freedom and non-centrality parameter $\lambda=T\frac{\mu_i^2}{\sigma^2}$. But what can we say about the distribution of the off-diagonal elements $X'X_{i,j}$?

Let $x_i\in\mathbb{R}^N$ be multivariate normal distributed with mean vector $\mu\in\mathbb{R}^N$ and no correlation: $x_i\sim N(\mu,\sigma^2 I_N)$. Given $T$ iid samples, define the matrix $$X:=\left( \begin{array}{c} x_1' \\ \vdots \\ x_T'\end{array} \right)\in\mathbb{R}^{T \times N}$$ I am interested how the resulting distribution of the elements of the matrix $X'X$ can be characterized. Clearly, the diagonal element $X'X_{i,i}$ contain the sum of independent squared normal variables with mean vector $\mu_i$ and variance $\sigma^2$. Therefore $\sigma^2X'X_{i,i}$ corresponds to a non-central chi squared distribution with $T$ degrees of freedom and non-centrality parameter $\lambda=T\frac{\mu_i^2}{\sigma^2}$. But what can we say about the distribution of the off-diagonal elements $X'X_{i,j}$?

Update: For a better understanding I appreciate every comment that helps to solve the simplified problem with $\mu=0$.