Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ be a $C^k$ function that is not compactly supported.
Is it always possible to find a diffeomorphism $\phi$, such that $\phi\circ f$ is compactly supported (and is it still $C^k$)? If not, how about a homeomorphism?
(You can choose $k$ to be as large as you need, in case that helps.)
Let $n = m = 1$ and consider $f(x) = x$. Suppose $\phi$ is a bijection $\phi : \mathbb{R} \to \mathbb{R}$ s.t. $\phi \circ f$ is compactly supported. Then in particular, there are at least 2 points $x,y$ s.t. $x \neq y$ and $\phi\circ f(x) = \phi \circ f (y) = 0$. Then $f(x) = \phi^{-1}(0) = f(y)$. But $f$ is an injection, so $x =y$, a contradiction.
So no, this doesn't always hold (note every diffeomorphism is a bijection, as well as every homeomorphism).