Double "orthogonal" group structure on a space. I am trying to find an appropriate setting for describing the following observation.
When studying ${\Bbb R}^3$ we observe a group operation $\omega\colon ({\Bbb R}^+,\cdot) \times {\Bbb R}^3 \to {\Bbb R}^3$ where the orbits are rays and as an orbital basis (i.e. a set consisting of exactly one point from every orbit) we can chose ${\Bbb S}^2$ or a deformed or scaled version thereof.
We can also observe a group action $\alpha\colon SO(3) \times {\Bbb R}^3 \to {\Bbb R}^3$ where the orbits look like ${\Bbb S}^2$ and as an orbital basis we can chose rays.
We thus have two groups operating on ${\Bbb R}^3$ which are, in a very lose sense, dual or orthogonal to each other. The orbits of the one group can form an orbital basis of the other and vice versa. And, whenever I am talking of "deformations" again the other group somehow enters the scene.
There are other, similar examples of a more trivial nature. I can understand ${\Bbb R}^3$ as a bundle of planes or as a bundle of lines. Again we can phrase this in similar terms. We have two groups, given by the additive groups of the one and the two dimensional vector space and on we go.
I am aware that this is a terrible form of a question (and I apologize for that) since I not yet really know which structure I am looking for exactly. Probably I am looking for some kind of bundle structure or fibre structure, but there I only have one group. What I am looking for is the possibility to have two (more?) groups in place with the original space (in the example ${\Bbb R}^3$) admitting several different group structures to enter the scene.
 A: I have the impression that your observation is based on your bundle being already a direct product, and each factor already is a Lie group structure in its own right.
So in your example (by the way, you should remove the origin to get a simply transitive action) you have $P:=\mathbb{R}^3\backslash\{0\}\cong\mathbb{S}^2\times\mathbb{R}_{>0}$ and both the maps
$$
x\mapsto\|x\|~~~~~\text{and}~~~~~x\mapsto\frac{x}{\|x\|}
$$
may serve as basis projection to turn $P$ into a principal bundle over the respective factor.
However, usually the structure of a manifold being a vector/principal/whatever bundle over some other manifold becomes fruitful and non-trivial if the bundle is non-trivial, otherwise you usually end up formulating stuff inside a product manifold way more complicated than necessary. And, as I said above, the appropriate setting you are looking for seems to be just a trivial bundle where the basis space is also a Lie group, making it possible to consider either factor as basis and as fiber, respectively. You can of course replace "trivial" by "trivializable" to cover your example of deformed spheres.
