= Solution
Put $S_1=\sum_{i=1}^nX_i$, $S_2=\sum_{i=1}^nX_i^2$. The joint <normal distribution> density can be written as
$$
p_\mu(\boldsymbol x)=(2\pi)^{-n/2}e^{-n/2}\mu^{-n}\exp\left(-\frac{S_2}{2\mu^2}+\frac{S_1}{\mu}\right).
$$
The <Fisher-Neyman factorization theorem> says that, for a family dominated by a common measure, a statistic $T$ is a <sufficient statistic> when its densities have the form $g_\mu(T(\boldsymbol x))h(\boldsymbol x)$ with $h$ independent of the parameter. The displayed factorization therefore shows that \b[$(S_1,S_2)$ is a two-dimensional <sufficient statistic>].
For <maximum likelihood estimation>, discard parameter-independent terms from the <log-likelihood>:
$$
\ell(\mu)=-n\log\mu-\frac{S_2}{2\mu^2}+\frac{S_1}{\mu},\qquad
\ell'(\mu)=\frac{S_2-S_1\mu-n\mu^2}{\mu^3}.
$$
If $S_2>0$, the quadratic equation has exactly one positive root. The numerator is positive before that root and negative after it, while the <likelihood> tends to zero at both ends of $\mu>0$. Hence the <normal likelihood with variance equal to squared mean> gives the global maximizer
$$
\boxed{\widehat\mu=\frac{\sqrt{S_1^2+4nS_2}-S_1}{2n}}.
$$
This holds almost surely. The exceptional all-zero sample has $S_2=0$ and <likelihood> proportional to $\mu^{-n}$, so its supremum is approached as $\mu\downarrow0$ and there is no positive maximizer.
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