TITLE: Which continuous functions are polynomials?
QUESTION [17 upvotes]: I posted this on the new math.SE website but didn't get much of a response, so I am reposting it here.
Suppose $f$ is a continuous $\mathbb{R}$-valued function on $\mathbb{R}^n$.  What type of conditions on $f$ guarantee it is a polynomial up to homeomorphism. That is, when can I find a homeomorphism $\phi:\mathbb{R}^n \to \mathbb{R}^n$ such that $\phi^* f = f \circ \phi \in \mathbb{R}[x_1,\ldots, x_n]$?
Some related questions:

A necessary condition in the case of $n = 1$ is that point inverse images of $f$ must be finite (since a polynomial has only finitely many roots).  Is this sufficient?
What if we replace $\mathbb{R}$ by $\mathbb{C}$?  
What if we look at smooth functions and diffeomorphism instead?  (I tried playing around with the implicit function theorem but didn't get anywhere).
What about the complex analytic case?

I'm not quite sure how to tag this, so feel free to edit them.

REPLY [4 votes]: I'm going to go out on a limb and make a partial conjecture based on Tom Goodwillie's comment.  A function $f:\mathbb{R}^n \to \mathbb{R}$ is topologically conjugate to a polynomial $p(x)$ only if it is topologically conjugate to a continuous function $q(x0$ of finite type which is nowhere constant.  By "finite type" I mean that there is a tiling of $\mathbb{R}^n$ by finitely many convex regions, such that the restriction of $q(x)$ to each region is non-constant and either linear or of the form $1/\phi(x)+c$, where $\phi$ is linear and $c$ is constant.
In this version of the answer, I'm going out on a limb for a second time.  I first conjectured that $q$ should simply be linear on each convex piece, and Richard Borcherds quickly found a counterexample to that in two variables.
I don't mean this to be a sufficient condition, since clearly it is not sufficient when $n=1$.  A "finite type" function in the above sense can be bounded, while a polynomial cannot be bounded.  Maybe it is a sufficient condition as a topological characterization of rational functions with no poles.  For polynomials specifically, there are strong restrictions on the behavior at infinity, but per Richard's example, they are somewhat looser than I first thought.
There is a relevant pair of results due to Whitney and Goresky.  Whitney proved that every analytic variety in $\mathbb{R}^n$ is a Whitney stratified space.  Goresky proved that every Whitney stratified space in $\mathbb{R}^n$ is supported on a piecewise smooth triangulation (but not necessarily one which is finite type).  It is easy to ride roughshod over subtleties as I already did, but these results seem like a good way to get started with the problem.
The smooth and complex cases of the problem seem more complicated for various reasons.