Decidability of bitstring languages with no ones' complements For any string $s \in \{0,1\}^*$, let the ones' complement of $s$, denoted $1_c(s)$, be the string obtained by flipping all bits in $s$. For example, $1_c(001100) = 110011$. Now call a language $L \subset \{0,1\}^*$ ones' complement asymmetric or OCA  for short, if for every string $s \in \{0,1\}^+$, precisely one of $s$ and $1_c(s)$ is in $L$.

I want to come up with a mapping reduction from a language $L \subset \{0,1\}^*$ to an OCA language (and vice-versa) so as to prove the existence of decidable/recognizable and undecidable/co-recognizable and undecidable, etc., OCA languages. My problem is especially that we require every string in $\{0,1\}^+$ to be such that either it or its ones' complement appears in any OCA language. This seems particularly problematic if $L$ lacks strings of a particular length, meaning it isn't obvious how we'd go about constructing an OCA language from $L$.
I don't think there exist recognizable-but-undecidable OCA languages, because if we have a Turing machine M that recognizes a given OCA language, then we can create a derived Turing machine that incrementally simulates M on both its input and its one's-complement (flipping back and forth, trying a finite number of steps in each simulation) until one input or the other has been recognized.

(Note: the derived Turing machine will also need to be given specific logic to decide the empty string; but that's obviously possible to do for any given language.)

Therefore, you can't construct any reduction mapping that would let you prove the existence of recognizable-but-undecidable OCA languages.