Equivalence of independent set and set packing
The exact meaning of "equivalent" isn't obvious but you have shown something deeper than the normal equivalence under reductions considered for NP-complete problems.

You've demonstrated what's known as a parsimonious reduction between the two problems.  Ordinarily, reductions between NP-complete problems are many-one reductions: they only have the property that problem&nbsp;A has a solution if, and only if, problem&nbsp;B has a solution. For example, when you reduce 3SAT to 3-Colourability, you produce a graph&nbsp;$G$ that is 3-colourable if, and only if, the original formula&nbsp;$\varphi$ is satisfiable. However, if $\varphi$&nbsp;has $N$&nbsp;variables, the number of satisfying assignments could be anything between zero and&nbsp;$2^N$, inclusive, whereas the number of 3-colourings of any graph is a multiple of six because of permutations of the set of colours.

The point about parsimonious reductions is that they're one-to-one. Your reduction sets up a bijection between solutions of the independent set problem and solutions of the corresponding set packing problem. Parsimonious reductions are useful because they preserve optimization and (approximate) counting versions of the problem. So your reduction also shows that the problem of finding the biggest independent set is as hard as finding the set packing using the most sets and that the problem of counting all the independent sets is as hard as to counting all the set packings.

There is a wider class of reductions that also preserve counting and approximate counting. These are the approximation-preserving reductions of Dyer et al.&nbsp;[1]. These are oracle reductions and relax the one-to-one requirement of parsimonious reductions to what is, essentially, "If you know (an approximation of) one, you can easily compute (an approximation of) the other." In particular, AP reductions can easily deal with the factor of&nbsp;$q!$ that's inherent in any reduction to the $q$-colouring problem.  As the name suggests, AP reductions preserve approximability, in the sense that, if there's an AP-reduction from A to&nbsp;B and there's an FPRAS for&nbsp;B, then there's an FPRAS for&nbsp;A, too.



[1] Dyer, Goldberg, Greenhill, Jerrum. The relative complexity of approximate counting problems. Algorithmica 38(3):471&ndash;500, 2003. DOI; free PDF