An Auto-correlated Cross-section?

When there is a time-series $\left\{X_1,\cdots,X_T\right\}$, one can compute sample auto-correlations using the data. For example, $\widehat{\mathrm{Cov}}\left[X_t,X_{t-1}\right]=\frac{1}{T-1}\sum_{t=2}^T{\left(X_t-\bar{X}\right)\left(X_{t-1}-\bar{X}\right)}$ is the sample first-order auto-covariance and the sample first-order auto-correlation is $\widehat{\mathrm{Corr}}\left[X_t,X_{t-1}\right]=\widehat{\mathrm{Cov}}\left[X_t,X_{t-1}\right]/\widehat{\mathrm{Var}}\left[X_t\right]$.

In a cross-section $\left\{X_1,\cdots,X_N\right\}$, there is no serial order among observations; hence, there is no order among the auto-correlations as well (i.e. $\mathrm{Corr}\left[X_i,X_j\right]=\rho,\forall{}i\ne{}j$). Is it possible to estimate the common auto-correlation $\rho$ using one cross-section here?

For instance, the sample mean is $\bar{X}=\hat{\mathrm{E}}\left[X_i\right]=\frac{1}{N}\sum_{i=1}^{N}{X_i}$ and the sample variance is $s^2=\widehat{\mathrm{Var}}\left[X_i\right]=\frac{1}{N-1}\sum_{i=1}^N{\left(X_i-\bar{X}\right)^2}$, respectively. The population mean and covariance may be $\mathrm{E}\left[\begin{pmatrix}X_1\\X_2\\\vdots\\X_N\end{pmatrix}\right]=\begin{pmatrix}\mu\\\mu\\\vdots\\\mu\end{pmatrix}$ and $\boldsymbol{\Sigma}=\mathrm{Cov}\left[\begin{pmatrix}X_1\\X_2\\\vdots\\X_N\end{pmatrix}\right]=\begin{pmatrix}\sigma^2&\rho\sigma^2&\cdots&\rho\sigma^2\\\rho\sigma^2&\sigma^2&\cdots&\rho\sigma^2\\\vdots&\vdots&\ddots&\vdots\\\rho\sigma^2&\rho\sigma^2&\cdots&\sigma^2\end{pmatrix}$.

I tried both ML (using a normal distribution) and MM first; for example, $\hat{\mathrm{E}}\left[X_iX_j\right]=\frac{1}{\binom{N}{2}}\sum_{\forall{}i\ne{}j}X_iX_j$ for MM and $\{\hat{\mu},\hat{\sigma},\hat{\rho}\}=\underset{\{\mu,\sigma,\rho\}}{\mathrm{argmax}}\ln{f\left(X_1,\cdots,X_N;\mu,\sigma,\rho\right)}$ for ML, where $\ln{f(\mathbf{X};\boldsymbol{\theta})}=-\frac{N}{2}\ln{(2\pi)}-\frac{1}{2}\ln{|\boldsymbol{\Sigma}|}-\frac{1}{2}(\mathbf{X}-\boldsymbol{\mu})^\top\boldsymbol{\Sigma}^{-1}(\mathbf{X}-\boldsymbol{\mu})$. However, it seems neither ML nor MM works.

Once the metric space is identified (on a knowledge-driven fashion), you can create your spatial weight matrix and use it to derive $\boldsymbol{\Sigma}$.