Are all decidable languages mapping reducible to its complement?
Let $L_1,L_2$ be two languages over $\{0,1\}$. A (computable) mapping reduction from $L_1$ to $L_2$ is a computable function $f\colon \{0,1\}^* \to \{0,1\}^*$ such that for all $x \in \{0,1\}^*$ it holds that $x \in L_1$ iff $f(x) \in L_2$.

In your example, $L_1 = \{0\}$ and $L_2 = \overline{\{0\}}$, there does exist a mapping reduction, given by
$$
f(x) = \begin{cases} 1 &amp; \text{if }x = 0, \\ 0 &amp; \text{if }x \neq 0.\end{cases}
$$
Clearly $f$ is computable, and you can check that $x \in L_1$ iff $f(x) \in L_2$.

Hence your example doesn't work. You can check, however, that $\emptyset$ and $\{0,1\}^*$ are two examples of languages not reducible to their complement. Indeed, these are the only examples, which you can try to prove.