Assume we have a homeomoprhism $\phi:M\rightarrow M$, where $M$ is a topological manifold which admits at least one smooth structure.


Is it always possible to construct a smooth structure on $M$ w.r.t to it $\phi$ will be a diffeomorphism?


Of course when there is little freedom in defining the smooth structure the answer is clearly no in general: Take the classic example of $M=\mathbb{R}$, $\phi(x)= x^3$. $\phi$ is a homeomorphism, but its inverse is not smooth w.r.t to the standard smooth strucutre on $\mathbb{R}$. (which is the only one that exists on $\mathbb{R}$). Note that $\phi$ itself is smooth.


**Update:** As pointed out by Andy Putman I was wrong to think there is no smooth structure making $\phi$ a diffeomorphism. (Such a structure in fact exsits, is clearly different from the standard one, although diffeomorphic to it).


I guess in most cases the answer will be negative. However, I would still like to find some sufficient or necessary conditions on $M$ and $\phi$ that ensure such a smooth structure exists. (In particular , is it always possible to ensure that at least one of $\phi$,$\phi^{-1}$ is smooth?).



When a homeomorphism is a diffeomorphism w.r.t to a suitable smooth structure?
Let me first answer your last question in the negative: there exist homeomorphisms $f:M \rightarrow M$ of smoothable manifolds $M$ such that neither $f$ nor $f^{-1}$ are smooth with respect to any smooth structure on $M$. In fact, we can take $M = S^3$. Bing has constructed homeomorphisms $f:S^3 \rightarrow S^3$ such that $f \circ f = \text{id}$ and such that the fixed set of $f$ is the Alexander horned sphere. See


Bing, R. H.
A homeomorphism between the 3-sphere and the sum of two solid horned spheres. 
Ann. of Math. (2) 56, (1952). 354–362. 


The map $f$ (and hence the map $f^{-1} = f$) is not smooth with respect to any smooth structure on $S^3$ because of a theorem of P. Smith that says that the fixed set of any periodic diffeomorphism of $S^3$ is a smoothly embedded surface or circle. This is the beginning of a long story that culminated in the proof of the Smith conjecture, which says that any periodic diffeomorphism of $S^3$ is smoothly conjugate to an element of the orthogonal group. This was one of the early triumphs of Thurston's work; for a detailed account of it, see the book


The Smith conjecture. 
Papers presented at the symposium held at Columbia University, New York, 1979. Edited by John W. Morgan and Hyman Bass. Pure and Applied Mathematics, 112. Academic Press, Inc., Orlando, FL, 1984. xv+243 pp. ISBN: 0-12-506980-4


This brings me to my second point. It is **FALSE** that the map $g:\mathbb{R} \rightarrow \mathbb{R}$ defined via $g(x) = x^3$ is not a diffeomorphism with respect to any smooth structure on $\mathbb{R}$. It is true that $\mathbb{R}$ has a unique smooth structure (as does $S^3$ above), but this means something weaker than what you are claiming it means. Namely, if $X$ and $Y$ are any two smoothings of $\mathbb{R}$, then there exists a diffeomorphism $X \rightarrow Y$; however, this diffeomorphism might not be the identity.


Here is a sketch of how you construct the appropriate smooth structure. Constructing this smooth structure is equivalent to constructing a homeomorphism $\psi:\mathbb{R} \rightarrow \mathbb{R}$ such that $\psi \circ g \circ \psi^{-1}$ is a smooth map; the desired smooth structure is then obtained by pulling back the standard smooth structure on $\mathbb{R}$ via $\psi$. Here are some observations about $g$:


1. It has three fixed points, namely $-1$ and $0$ and $1$.
2. On the interval $(-\infty,-1)$, it shifts points to the left.
3. On the interval $(-1,0)$, it shifts points to the right.
4. On the interval $(0,1)$, it shifts points to the left.
5. On the interval $(1,\infty)$, it shifts points to the right.


It is easy to construct a diffeomorphism $h: \mathbb{R} \rightarrow \mathbb{R}$ with the above properties (since $0$ will be an attracting fixed point of $h$, the derivative of $h$ at $0$ will be a number $\alpha$ satisfying $0 < \alpha < 1$). But it is then an easy exercise to see that $h$ will be topologically conjugate to $g$.


