= Solution
For $\sigma>0$, the density is
$$
p_\sigma(x)=\frac1{\sqrt{2\pi}\sigma}
\exp\left(-\frac{x^2}{2\sigma^2}\right)
=g_\sigma(|x|),
$$
so the <Fisher-Neyman factorization theorem> proves that $|X|$ is sufficient.
Moreover,
$$
\frac{p_\sigma(x)}{p_\sigma(y)}
=\exp\left(-\frac{x^2-y^2}{2\sigma^2}\right)
$$
is independent of $\sigma$ exactly when $x^2=y^2$, equivalently $|x|=|y|$. The <likelihood-ratio criterion for minimal sufficiency> therefore proves that
$$
\boxed{|X|\text{ is minimal sufficient}.}
$$
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