I will answer in the negative the question "If $f:I\to I$ is a strictly increasing absolutely continuous function, does $f^{-1}$ send Lebesgue null sets to Lebesgue null sets?"

Indeed, suppose that $f$ is an absolutely continuous strictly increasing bijection on $I$, such that all preimages of nullsets are null. Write $g:=f^{-1}$. Note that, under the hypotheses, $f$ and $g$ are homeomorphisms. Additionally, $m(f^{-1}(A))=m(g(A))$ is null for all $A\subseteq I$ null, so $g$ maps null sets to null sets, i.e. $g$ satisfies the Luzin property.

It appears to be a classical fact that a function is a.c. if and only if it is continuous, BV, and has the Luzin property. Since $g$ is continuous and has the Luzin property, and is a bounded increasing function, it follows that $g$ is a.c.

Thus, $f$ has to also have the property that its inverse is a.c. On the other hand, strictly increasing a.c. functions need not have a.c. inverses; indeed, if $g=c(x)+x$ and $f=g^{-1}$, then $g$ is clearly not a.c., but $f$ is Lipschitz (hence a.c.) and strictly increasing.

As suggested in the preceding, the correct condition has something to do with $g=f^{-1}$ being absolutely continuous, which might be thought of as $f'$ being not too small (i.e. $\frac{1}{f'}$ being integrable).

I will answer in the negative the question "If $f:I\to I$ is a strictly increasing absolutely continuous function, does $f^{-1}$ send Lebesgue null sets to Lebesgue null sets?"

Indeed, suppose that $f$ is an absolutely continuous strictly increasing bijection on $I$, such that all preimages of nullsets are null. Write $g:=f^{-1}$. Note that, under the hypotheses, $f$ and $g$ are homeomorphisms. Additionally, $m(f^{-1}(A))=m(g(A))$ is null for all $A\subseteq I$ null, so $g$ maps null sets to null sets, i.e. $g$ satisfies the Luzin property.

It is a classical fact that a function is a.c. if and only if it is continuous, BV, and has the Luzin property. Since $g$ is continuous and has the Luzin property, and is a bounded increasing function, it follows that $g$ is a.c.

Thus, $f$ has to also have the property that its inverse is a.c. On the other hand, strictly increasing a.c. functions need not have a.c. inverses; indeed, if $g=c(x)+x$ with $c:[0,1]\to[0,1]$ the Cantor function, and $f=g^{-1}$, then $g$ is clearly not a.c., but $f$ is Lipschitz (hence a.c.) and strictly increasing.

As suggested in the preceding, the correct condition has something to do with $g=f^{-1}$ being absolutely continuous, which might be thought of as $f'$ being not too small (i.e. $\frac{1}{f'}$ being integrable).