There are strictly increasing $C^1$ functions that map sets of positive measure to sets of measure zero. Here is a construction:
Let $C\subset [0,1]$ be a Cantor set of positive measure. For a construction, see https://en.wikipedia.org/wiki/Smith-Volterra-Cantor_set. Let $g(x)=\operatorname{dist}(x,C)$. The function $g$ is clearly continuous and equal zero on $C$. In fact $g$ is a $1$-Lipschitz function. Let
$$
f(x)=\int_0^x g(t)\, dt.
$$
The function $f$ is $C^1$ and it is strictly increasing. Indeed, if $y>x$, then
$$
f(y)-f(x)=\int_x^y g(t)\, dx>0
$$
because the interval $[x,y]$ is not contained in the Cantor set $C$ and therefore it contains an interval where $g$ is positive.
On the other hand $f'=g=0$ on $C$ which has positive measure and $f(C)$ has measure zero since $m(f(C))=\int_C f'(t)\, dt=\int_C g(t)\, dt=0$.
As was pointed out by Mateusz Kwaśnicki in his comment, this construction gives the following result:
Theorem. Let $f$ be as above. Then there is a Riemann integrable function $h$ such that $h\circ f$ is not Riemann integrable.
Proof. The set $f(C)$ is homeomorphic to the Cantor set ($f$ is strictly increasing so it is a homeomorphism) and has measure zero as explained above. Let
$$
h(x)=\begin{cases}
1 & \text{if $x\in f(C)$}\\
0 & \text{if $x\not\in f(C)$.}
\end{cases}
$$
The function $h$ is Riemann integrable with the integral equal zero since it is bounded and continuous outside the set $f(C)$ of measure zero (because $\mathbb{R}\setminus f(C)$ is open and $h=0$ there). However,
$$
(h\circ f)(x)=\begin{cases}
1 & \text{if $x\in C$}\\
0 & \text{if $x\not\in C$.}
\end{cases}
$$
is not Riemann integrable since it is discontinuous on a set $C$ of positive measure. $\Box$
