= Solution
The <independent head-tail reversal symmetry> means that the squared <dot product> interaction <energy> is unchanged when any individual $\mathbf s_i$ is replaced by $-\mathbf s_i$. All configurations $\mathbf s_i=\sigma_i\mathbf n$, with $\sigma_i=\pm1$ independently, attain the aligned ground-state pair <energy>. There is consequently no energetic preference for either end of a molecular axis. Averaging over these equally weighted signs gives $\langle\mathbf s_i\rangle=0$ and \b[$\langle\mathbf m\rangle=0$ even in the nematically ordered phase]. The surviving order is the unoriented axis, represented by a <nematic director> with $\mathbf n\sim-\mathbf n$. A second-moment tensor can detect this order while a polar <vector> cannot.<Lectures on Landau Theory of Phase Transitions by Peter D. Olmsted>[§4.3, printed pp.10–11, head-tail equivalence and the tensor order parameter]{ref}
The word “always” needs the <statistical ensemble> or apolar interpretation: it is not an identity for each finite configuration. For example, choosing all $\sigma_i=+1$ is also a ground-state representative and gives $\mathbf m=\mathbf n$. In an unbiased sign <statistical ensemble> the typical residual polar <vector> has size of order $N^{-1/2}$, because $\langle|\mathbf m|^2\rangle=1/N$ for these independent ground-state signs. This does not make polar order an equilibrium <nematic order parameter>. The tensor below is unchanged by each sign flip and thus removes the arbitrary labeling of rod ends.
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