I was wondering about this question, but I simply don't have the tools to tackle it.
Suppose we have some function $f:\mathbb{R} \to \mathbb{R}$ that is monotonic at least in some interval $(a,b)$. Now consider a second function $g: \mathbb{R}\to\mathbb{R}$ such that $f(x) \neq g(x) \ \forall \ x \in (a,b)$, but there exists a single $y \neq x \in (a,b)$ such that $f(x) = g(y)$. In other words, the image of $g$ coincides with the image of $f$ for the subset $(a,b) \subset \mathbb{R}$, but $f(x)$ never maps to the same element as $g(x)$, so $g$ is a "shuffling" of $f$, in a sense, with no fixed points.
Under the conditions above:
Can we establish some $g$, by proof or example, such that it is not Riemann and/or Lebesgue integrable in the interval $(a,b)$? I suspect that the answer is yes for the Riemann case. No clue for the Lebesgue case.
If $g$ is integrable, does $$ \int_a^b f(x)\,\mathbb{d}x = \int_a^b g(x)\,\mathbb{d}x $$ always follows?
You can easily get examples with $g$ non-measurable, which precludes it from being integrable. For instance take $(a,b)=(0,1)$, and $f(x)=x$. Let $E\subset(0,1)$ be any uncountable non-measurable set and $h:E\to (0,1)$ a bijection. Then take $$ g(x)=h(x)\,1_E. $$ Then $g$ is not measurable because $g^{-1}(0,1)=E$, while $g(0,1)=(0,1)$. One can easily tweak $h$ so that $g$ agrees with $f$ at some points.
The second condition fails very easily. An integral takes into account where the function takes its values. Again take $f(x)=x$, and $g(x)=2(x-x^2)$. Then $$ \int_0^1f=\frac12\,\qquad\qquad\int_0^1 g=\frac13. $$ We have $f(1/2)=g(1/2)$.