This was motivated by the definition of a smooth embedding as an injective smooth immersion that is a topological embedding; I see no reason a-priori for the image of an embedding as defined to be diffeomorphic to the original manifold unless this was true (and if they aren't diffeomorphic then calling it a smooth embedding doesn't seem to make sense to me).
For example, it is well-known that the $x \mapsto x^3$ is a smooth homeomorphism that is not a diffeomorphism. However, its image is clearly diffeomorPHIC to its domain (both being $\mathbb{R}$ with the standard smooth structure). My question is therefore this: is it true in general that if $f: M \rightarrow N$ is a smooth homeomorphism then $M$ and $N$ are diffeomorphic?
EDIT: the key thing I somehow missed was the "immersion" part, yes a smooth homeomorphic IMMERSION is a diffeomorphism by the Inverse Function Theorem. But my question still stands.
While I am quite sure I had seen this question before on MSE (and that I even answered it), after 20 minutes searching I could not find a duplicate. Maybe it was removed. So, here is an answer.
Proposition. Suppose that $X, Y$ are smooth 7-dimensional manifolds homeomorphic to $S^7$. Then there exists a smooth homeomorphism $X\to Y$.
At the same time, there are 28 diffeomorphism classes of such manifolds (Milnor's exotic spheres). Hence, you get examples of smooth homeomorphisms between non-diffeomorphic manifolds.
Let's prove the proposition. Milnor proved that all manifolds $X, Y$ as above are obtained by gluing two $7$-dimensional balls via a diffeomorphism of their boundaries. Thus, we only need to prove:
Lemma. Let $f: S^6\to S^6$ be a diffeomorphism. Then $f$ extends to a smooth homeomorphism $F: B^7\to B^7$ of the closed 7-dimensional unit ball. Moreover $F$ can be chosen to be a diffeomorphism away from the origin.
Proof. Let $\phi(r), r\in [0,1]$, be a smooth strictly monotonic function such that $\phi(0)=0, \phi(1)=1$ and the derivatives of all orders of $\phi$ vanish at $0$ and $\phi'(r)\ne 0$ for all $r\ne 0$. I will be using the polar coordinates $(r, \theta)$ on $B^7$, where $r\in [0,1]$, $\theta\in S^6$. Then the desired extension of $f$ is given by the formula $$ F(r,\theta)=(\phi(r), f(\theta)). $$
qed
Edit. I found an answer by Greg Kuperberg here to a similar MO question. Greg claims even more, namely that if two smooth manifolds are PL-equivalent then there is a smooth homeomorphisms between them. However, I could not follow his argument.