Given any smooth manifold $(M,\mathscr{A})$ with a specified smooth structure $\mathscr{A}$, we can identify uncountably many distinct smooth structures $(\mathscr{B}_s)_{s \ge 0}$ such that $(M,\mathscr{B}_s)$ is also a smooth manifold.
So how do we go about choosing a standard smooth structure to work with and do calculations with? How can we justify any of the uncountably many choices as the "best" one?
Context: My confusion is a result of solving the following problem (i.e. my solution raises more questions than it answers) from Lee's Introduction to Smooth Manifolds, 1-6 on p.30:
Let $M$ be a nonempty topological manifold of dimension $n \ge 1$. If $M$ has a smooth structure, show that it has uncountably many distinct ones. [Hint: first show that for any $s>0$, $F_s(x)=|x|^{s-1}x$ defines a homeomorphism from $\mathbb{B}^n$ to itself, which is a diffeomorphism if and only if $s=1$.]
This answer and this question both seem to imply that all of these smooth structures should be diffeomorphic, but if they were diffeomorphic, then wouldn't the smoothness of transition maps between coordinate charts imply that they are equal (since any smoothness structure is a maximal atlas)? Anyway, even if they were all diffeomorphic, that still doesn't resolve the issue of which one should be the "standard" one and which one to choose for calculations, etc.
For example, on p. 40 of this same book, it says that:
...$\mathbb{R}^4$ has uncountably many distinct smooth structures, no two of which are diffeomorphic to each other! The existence of nonstandard smooth structures on $\mathbb{R}^4$ (called fake $\mathbb{R}^4$'s) was first proved by Simon Donaldson and Michael Freedman in 1984 as a consequence of their work on the geometry and topology of compact 4-manifolds...
So when working with $\mathbb{R}^4$, how does one decide which smooth structure is the "standard" smooth structure, and how can someone verify that they are working with the correct smooth structure?
My attempt: Does the answer have to do with the fact that any topological manifold already comes pre-equipped with a collection of charts which are appropriate homeomorphisms but not necessarily smoothly compatible with each other, and thus any smooth structure is a strict subset of the family of charts inherited from the topological manifold structure? Such that any two distinct smooth structures are both strict subsets of the family of charts from the underlying topological manifold? So they are necessarily equivalent up to homeomorphism? I still don't see how to show that they are equivalent up to diffeomorphism. Also the fact that any smooth structure on a set induces a topological manifold structure on the same set would seem to suggest that any smooth atlas is not strictly contained in the family of charts resulting only from the topology of the underlying space, although I am not sure either way.