Any countable Borel equivalence relation can arise from a Borel action of the free group on $\omega$ generators, which is a natural enough group. You should probably ask instead whether the action can be natural. Can the action be free, for example, as Joel discusses.

The field of countable Borel equivalence relations has gained a tremendous amount from studying equivalence relations arising from natural group actions. However, for the case of Turing equivalence, viewing this relation as arising from the Borel action of a countable group does not appear to be a helpful way of thinking.

You might be interesting in the following open question (from my thesis). The question essentially asks whether the Turing reductions are the unique way to generate Turing equivalence:

Suppose that $\{\psi_i\}_{i \in \omega}$ is any countable collection of partial Borel functions closed under composition such that $x$ and $y$ are Turing equivalent if and only if there exists an $i$ and a $j$ such that $\psi_i(x) = y$ and $\psi_j(y) = x$. Suppose also that $\{\varphi_i\}_{i \in \omega}$ is an enumeration of the Turing reductions. Then must there be a pointed perfect set $P$ and a function $u: \omega^2 \rightarrow \omega^2$ such that for every pair of Turing reductions $\varphi_i$ and $\varphi_j$, if $x,y \in P$ and $\varphi_i(x) = y$ and $\varphi_j(y) = x$, then $\psi_k(x) = y$ and $\psi_l(y) = x$, where $u(i,j) = (k,l)$?

It is probably optimistic to hope that the answer to the above question is "yes", when we have no idea how to approach this problem. However, beyond the question being nice in and of itself, a positive answer would have a large number of consequences. It would imply Martin's conjecture on Turing invariant functions is true, for example, which in turn would have a lot of consequences for the theory of countable Borel equivalence relations.

A positive answer would also imply any universal countable Borel equivalence relation must be able to witness its universality uniformly, regardless of the way it is generated. A nice implication of this would be that any increasing union of non-universal countable Borel equivalence relations must be non-universal.