Another Smooth Structure on $\mathbb R$, clarification needed, John M. Lee In Introduction to Smooth Manifolds by John M. Lee, we have two statements, I hope I got the situation correct
The first is on p. 17,

Example 1.23 (Another Smooth Structure on $\mathbb R$). Consider the homeomorphism $\psi: \mathbb R \to \mathbb R$ given by
$$
\psi(x) = x^3 \tag{1.1}
$$
The atlas consisting of the single chart $(\mathbb R,\psi)$ deﬁnes a smooth structure on $\mathbb R$. This chart is not smoothly compatible with the standard smooth structure, because the transition map $\text{Id}_\mathbb R \circ \psi^{-1} (y)=  y^{1/3}$ is not smooth at the origin. Therefore, the smooth structure deﬁned on $\mathbb R$ by $\psi$ is not the same as the standard one. ...//

While on page 40,

In fact, as you will see later, there is only one smooth structure on $\mathbb R$ up to diffeo-
morphism (see Problem 15-13). More precisely, if $\mathcal A_1$ and $\mathcal A_2$ are any two smooth structures on $\mathbb R$, there exists a diffeomorphism $F :(\mathbb R,\mathcal A_1)  \to (\mathbb R,\mathcal A_2)$. In fact, it follows from work of James Munkres [Mun60] and Edwin Moise [Moi77] that every topological manifold of dimension less than or equal to 3 has a smooth structure that is unique up to diffeomorphism. ...

So to make things clear, in our case, one has to find a diffeomorphism $F:(\mathbb R,\psi) \to (\mathbb R,\text{Id}_\mathbb R )$. If I got it correctly this map should be $F(x) = x^3$, since $\text{Id}_\mathbb R \circ F \circ \psi^{-1} = \text{Id}_\mathbb R$, correct ?
I'm confused about diffeomorphic smooth structures and the differentiabilities they define. So it seems that if there is a diffeomorphism between two smooth structures on the same manifold, it doesn't mean that every map or function that is smooth wrt one of these structures will be smooth wrt the other
 A: Your understanding of the situation is correct: $\mathbb R$ is endowed with two distinct smooth structures (the one induced by $\text{Id}_\mathbb R$ is the standard smooth structure). However, the smooth manifolds $R = (\mathbb R,\text{Id}_\mathbb R )$ and $R' = (\mathbb R,\psi)$ are diffeomorphic (via your diffeomorphism $F : R' \to R$).
As you observe, this does not mean that the sets $C^\infty(R,M)$ and $C^\infty(R',M)$ of smooth maps into a smooth manifold $M$ agree, not even for $M = R$. BUT: $F$ induces a bijection
$$F^* : C^\infty(R,M) \to C^\infty(R',M),\qquad F^*(g) = g \circ F .$$
If you think about it, you will find that this phenomemon is not that surprising. The general pattern is this:
You start with a certain category $\mathbf C$ (e.g. the category of topological spaces and continuous maps) and endow some of its objects $X$ with an additional structure $\mathcal S$ (e.g. a smooth structure). Write this as $(X,\mathcal S)$. Define morphisms $\phi : (X_1,\mathcal S_1) \to (X_2,\mathcal S_2)$ as morphisms $\phi : X_1 \to X_2$ in $\mathbf C$ which are "compatible with the additional strucures" in a suitable sense. This gives you an "enriched category"  $\mathbf D$ with a forgetful functor $V : \mathbf D \to  \mathbf C$.
The "enrichment process" for objects is in general not unique, i.e. the same basic object $X$ may receive distinct additional structures $\mathcal S, \mathcal S'$ which gives you distinct objects $(X,\mathcal S)$ and $(X,\mathcal S')$ of $\mathbf D$. These may be isomorphic or not. But even if they are isomomorphic, there is no reason to expect $Hom_{\mathbf D}(((X,\mathcal S), \mathscr Y)  = Hom_{\mathbf D}(((X,\mathcal S'), \mathscr Y)$ or $Hom_{\mathbf D}(\mathscr Y, (X,\mathcal S))  = Hom_{\mathbf D}(\mathscr Y, (X,\mathcal S'))$ for all objects $\mathscr Y$ of $\mathbf D$. This may be true in some cases, but in general we can only say that each isomorphism $\iota : (X,\mathcal S) \to (X,\mathcal S')$ induces natural bijections between these morphism sets.
Let us consider examples.

*

*Let $\mathbf C$ be the category of sets and functions. Sets can be endowed with group structures, but of course any set with more than one element can be given distinct group structures. Enrichment produces the category of groups and homomorphisms. Now consider $X = \{0,1\}$. It can be given two group strucutures: One in which $0$ is neutral and one in which $1$ is neutral. Call the resulting groups $F_2$ and $F'_2$. Then clearly $Hom(F_2, F_2) = \{id, 0\}$ and $Hom(F'_2, F_2) = \{\tau, 0\}$, where $0$ denotes the constant map with value $0$ and $\tau$ exchanges the points of $\{0,1\}$.


*Let $\mathbf C$ again be the category of sets and functions. Sets can be endowed with topologies, but any set with more than one element can be given distinct topologies. Enrichment produces the category of topological spaces and continuous maps. Now consider $\mathbb R$ as a set. It has a standard topology and we denote the corresponding space by $\mathbb R$. A new topology can be defined by requiring the bijection $b : \mathbb R \to \mathbb R, b(0) =1, b(1) = 0, b(x) = x$ else, to become a homeomorphism. Denote the resulting space by $\mathbb R'$. It is now clear that $Hom(\mathbb R, \mathbb R) \ne Hom(\mathbb R', \mathbb R)$; the identity is not contained in $Hom(\mathbb R', \mathbb R)$.
