Primitive Recursive Isomorphisms
Yes, people have looked at this and related groups, though not a whole lot as far as I can tell. For example:

Morozov, A. S. Turing reducibility as algebraic embeddability. Siberian Math. J. 38 (1997), no. 2, 312–313. Let $G_d$ denote the group of permutations of $\mathbb{N}$ computable below the Turing degree $d$. Morozov shows that $d_1 \leq_{T} d_2$ if and only if $G_{d_1} \hookrightarrow G_{d_2}$. (Note that a permutation and its inverse always have the same Turing degree. This no longer holds if we restrict to primitive recursive or poly-time reductions.)

Kent, C. F. Constructive analogues of the group of permutations of the natural numbers. Trans. Amer. Math. Soc. 104 (1962) 347–362. 

In fact, if I recall correctly, the primitive recursive permutations generate the group of all computable permutations. (If you restrict attention to the group of permutations $\pi$ such that both $\pi$ and $\pi^{-1}$ are primitive recursive, you get a smaller group.) Furthermore, every computable permutation can be written as a word of length six in the primitive recursive permutations and their inverses. I think the same result holds if you replace "primitive recursive" by "polynomial time." [I don't recall the reference just now, but I have a physical copy of the paper at home. I will post the reference when I return from traveling.]

Slightly different but related:

Combarro, E. F. Classification of subsets of natural numbers by computable permutations. Siberian Math. J. 45 (2004), no. 1, 125–135. 

The group of computable permutations is not finitely generated (Is that what you meant by "Are there generators to this group?" Every group has generators, for example, the entire group...). If it were, then there would be a c.e. set $I$ of indices such that the set $\{\varphi_i : i \in I\}$ was exactly the set of computable permutations (where $\varphi_i$ is the $i$-th computable function in some acceptable enumeration). It then takes a little argument to show that this cannot happen (start from the fact that the set of all indices of computable permutations is $\Pi^0_2$-complete).