The Microeconomics Lecture notes by Rubinstein has the following question in Problem set two.
Let $U, V: X \to \mathbb{R}$ be two utility representations of the preference relation $P$ (preference relations are a total ordering and a utility representation is a map that preserves this order). Is there always a strictly monotonic function $f: \mathbb{R} \to \mathbb{R}$ s.t. $V = f \circ U$. The tone in which the question is asked seems to suggest that there is not, but for my reasoning there should be:
One can define $\tilde{U}, \tilde{V}: \frac{X}{\sim} \to \mathbb{R}$ which are injective and then define $$\hat{f} = \tilde{V} \circ \tilde{U}^{-1}$$ on some subset $M$ of $\mathbb{R}$. $\hat{f}: M \to \mathbb{R}$ is strictly montonic and can be linearly interpolated in areas where $M$ is not dense and continuously completed in areas where $M$ is dense.
Edit: On a second thought the completion procedure is not completly trivial since the left and the right limit could differ. But you can just choose one consistently and it would still work I guess, $f$ just would not be continuous anymore.
Do I have an error in my reasoning here?
No there need not be such a function $f$, it is possible that you run out of space in some sense. Suppose $X=\mathbb N$ and the preference relation is just the usual ordering on $\mathbb N$. One way to represent this is just $V:\mathbb N\rightarrow \mathbb R$ the inclusion map. Another way to do this is $$U:\mathbb N\rightarrow \mathbb R,\ n\mapsto 1- \frac{1}{n+1}$$ Any map $f:\mathbb R\rightarrow\mathbb R$ with $V=f\circ U$ must map $1-\frac{1}{n+1}$ to $n$. But if $f$ should be strictly monotonic, it would have to send $1$ to a number which is larger then any $n\in\mathbb N$, which is impossible.
If you apply your procedure here, you could define $f$ for any $x<1$, but then get stuck.