Let $S \subset A^2$ and $S=\{(a,b): aSb\} .\ $Then $\ S^{-1}=\{(b,a):aSb\} \subset A^2.$ That means $S^{-1}$ is also equivalence relation, because every pair is in the same relation as in relation $S$ (so it has the same properties) but entries are switched.
This is my answer to the question but for me it feels kinda tottery, so I am seeking more algebraic explenation.
From the symmetry of the relation $S$, we can get $S^{-1}=S$. Thus equivalence of $S$ implies equivalence of $S^{-1}$.