If A is reducible to its complement, is the converse true as well?
In general whenever $A\le_m B$ via computable function $f$, we automatically have $\bar{A}\le_m \bar{B}$ via the same $f$. The reason is that $x\in A\iff f(x) \in B$ implies that $x \in \bar{A} \iff f(x) \in \bar{B}$.