= Solution
Use the quotient <lens spaces> and set $\zeta=e^{2\pi i/7}$. Let $g_j(z_1,z_2)=(\zeta z_1,\zeta^jz_2)$. The map
$$
F:S^3\to S^3,\qquad F(z_1,z_2)=(z_2,\overline{z_1})
$$
is a <homeomorphism> and satisfies
$$
F g_2(z_1,z_2)=(\zeta^2z_2,\zeta^{-1}\overline{z_1})
=g_3^2F(z_1,z_2),
$$
because $3\cdot2\equiv-1\pmod7$. Since $2$ is invertible modulo $7$, $g_3^2$ generates the entire target deck group. Hence $F$ maps cyclic orbits bijectively to cyclic orbits, as does its inverse, and descends to a <homeomorphism>:
$$
\boxed{L(7,2)\cong L(7,3).}
$$
This is an explicit map, so no classification theorem for <lens spaces> is required. It reverses <orientation>: complex conjugation in one coordinate reverses the ambient real <orientation>, while swapping the two complex coordinates preserves it.
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