= Solution
The <symmetry> is <electromagnetic reciprocity> applied to <mutual inductance>. For thin closed circuits in <vacuum>, the magnetic <vector> potential generated by circuit B is
$$
\mathbf A_B(\mathbf r)=\frac{\mu_0I_B}{4\pi}\oint_B\frac{d\boldsymbol\ell_B}{|\mathbf r-\mathbf r_B|}.
$$
Stokes theorem converts the <magnetic flux> through circuit A into its magnetic <vector> potential <circulation>. Dividing by $I_B$ gives <Neumann's mutual-inductance formula>,
$$
M_{AB}=\frac{\Phi_{A\leftarrow B}}{I_B}
=\frac{\mu_0}{4\pi}\oint_A\oint_B
\frac{d\boldsymbol\ell_A\cdot d\boldsymbol\ell_B}{|\mathbf r_A-\mathbf r_B|}.
$$
The <dot product> and separation are unchanged by interchanging A and B, so
$$
\boxed{M_{AB}=M_{BA}.}
$$
Each has units $\mathrm{Wb\,A^{-1}}=\mathrm H$. The winding orientations must be used consistently to determine the common sign. The same reciprocity holds in ordinary linear reciprocal magnetic media; it is not an unrestricted statement about biased nonreciprocal materials. Equivalently, a reversible <magnetic energy> function has equal mixed <derivatives> with respect to the two currents.
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