Note: $0$ is a natural number, and the first projection is $\pi_0$
Call a relation $R\subseteq\Bbb N^k$, $k>1$ nontrivial if and only if $\pi_i(R)$ is infinite for every $i<k$.
For any pair of nontrivial relations $A,B\subseteq \Bbb N^k$, $k>1$ we call a function $f:X\to Y$, where
$$\bigcup_{i<k}\pi_i(A)\subseteq X\subseteq\Bbb N$$
and
$$\bigcup_{i<k}\pi_i(B)\subseteq Y\subseteq\Bbb N,$$
a homomorphism from $A$ to $B$ if and only if $(x_0,\ldots,x_{k-1})\in A\implies(f(x_0),\ldots,f(x_{k-1}))\in B$. $f$ is an isomorphism if and only if it is an invertible homomorphism and $f(A)=B$.
What computability properties are preserved under...
...isomorphism?
...when $\operatorname{dom}(f)=A$?
...when $\operatorname{dom}(f)=\Bbb N$?
...computable isomorphism?
...when $\operatorname{dom}(f)=A$?
...when $\operatorname{dom}(f)=\Bbb N$?
...primitive-recursive isomorphism?
(primitive recursive implies total, so $\operatorname{dom}(f)=\Bbb N$, always)
For example:
If $A$ is non-R.E., and $f$ is primitive recursive, is $B$ non-R.E.?
If $A$ is recursive, and $f$ is noncomputable and total, is $B$ still recursive?
If $A$ is $\Pi^0_2$, and $f$ is [partial] recursive, is $B$ $\Pi^0_2$?
Trivially, when $k=1$ the answer to the general question is a resounding "no." Any infinite subset of $\Bbb N^1=\Bbb N$ is isomorphic to every other infinite subset of $\Bbb N$. Likewise, we can find a trivial relation whose relative computability is not preserved under isomorphism - pick a nice set $A_0$, and let $A=A_0\times\{0\}$; construct an isomorphism $f$ to send $A_0$ to the nastiest set you can think of - then $B$ is nasty. I would think that the general case is still a "no" even when $A$ is nontrivial, but I'm not sure. It seems reasonable that computable isomorphism preserves R.E.ness, but not necessarily recursiveness, and in general I imagine that an isomorphism of class $C$ preserves properties "as nice as $C$."