Let $U \subset \mathbb{R}^n$. Endow it with the standard smooth structure, given by smooth atlas $A_U=\{(U,id_U)\}$.
Then if one considers another homeomorphism $\phi:U \rightarrow U$, then is the smooth atlas $\{(U, \phi)\}$ somehow a subset of $A_U$? By the naive reasoning that $id_U$ is a "full chart" (maps everything $U \rightarrow U$).
Particularly, would this imply also that since $id_U$ is smooth, then $\phi$ is smooth?
You have two atlases $A_U$ and $A_\phi =\{(U, \phi)\}$ for the manifold $U$. Let us more generally consider a homeomorphism $\phi : U \to V$, where $V$ is an open subset of $\mathbb{R}^n$. If you like you can take $V = U$, but this does not make a conceptual difference.
If $\phi \ne id_U$, these atlases are not directly related, in particular $A_U \not\subseteq A_\phi$ $A_\phi \not\subseteq A_U$.
However, it is possible that both yield the same smooth structure (= maximal smooth atlas) on $U$. This is true if and only if the transition maps $\phi = \phi \circ id_U^{-1} : U \to V$ and $\phi^{-1} = id_U \circ \phi^{-1} : V \to U$ are smooth. The latter means that $\phi$ is a diffeomorphism $U \to V$ in the classical sense of multivariable calculus.
Not any homeomorphism is a diffeomorphism in that sense, for example $\phi : \mathbb{R} \to \mathbb{R}, \phi(x) = x^3$ is not.
If $V = U$, then the smooth manifolds $(U,A_U)$ and $(U,A_\phi)$ are diffeomorphic. In fact, $\phi : (U,A_\phi) \to (U,A_U)$ is a diffeomorphism of smooth manifolds (which is not the same as a diffeomorphism in the sense of multivariable calculus). This is true because with respect to the only two charts $id_U, \phi$ the map $\phi$ has coordinate representation $id_U \circ \phi \circ \phi^{-1} = id_U : U \to U$ and $\phi^{-1}$ has coordinate representation $\phi \circ \phi^{-1} \circ id_U^{-1} = id_U : U \to U$.