Let $A\subset \mathbb{R}^n$ a Borel set of finite Lebesgue measure. They define $A^*$ to be the ball centered at 0 with the same measure that $A$.
The symmetric-decreasing rearrangement of a measurable function $f:\mathbb{R}^n \to \mathbb{R}$ is then defined by
$$f^*(x):=\int_0^{\infty} \chi_{\{|f|>t\}^*}(x)dt,$$
by comparison to the "layercake" representation of $f$, namely $$f(x)=\int_0^{\infty} \chi_{\{f>t\}}(x)dt.$$
Note that one can equally define $f^*$ to be the radial symmetric, decreasing function such that the level sets of $f^*$ and $f$ have the same measure (or volume).
I don't know why, but it always occurs to me that the rearrangement process is a regularizing process.
Here's my question: If $f$ is $k$ times continuously differentiable, does it follow that $f^*$ possesses the same regularity?
In general the answer is no, it does not. A counter-example is the following. Given parabola $f$ (which is differentiable infinitely many times) on the interval $(-1,1)$, it has a symmetric decreasing rearrangement $f^*$ which is not differentiable at $0$. So for functions from $C^k$ we can guarantee just $PC^k$ (partially continuous of degree $k$). But if given function $f$ is monotone then, I think, your suggestion would be true.