Bayesian Epistemology, Bovens & Hartmann, OUP 2003, p. 22, appears to give this argument:
Let $f,g: (0,1) \rightarrow (0,1)$ be distinct functions, for which there's a $a \in (0,1)$ such that $f(a)=g(a)$.
Let $h^*: (0,1) \rightarrow (0,1)$ be $h^*(x)=1-x$, and consider $$f^*(x)=f(h^*(x)) \;, \\ g^*(x)=g(h^*(x)) \;.$$ If there's a $b \in (0,1)$ such that $f^*(b)=g^*(b)$, then for any continuous, strictly monotone increasing function $h$ from (0,1) onto (0,1), there is some value $c$ such that $f(h(c))=g(h(c))$.
The italicized passage is the proposition to be proven. That's the argument, which I hope I have represented correctly--the original wording is confusing. I see that the conclusion is justified: Since $h$ is onto $(0,1)$, there's a $c$ such that $h(c)=a$. The conclusion follows from this alone. (This $c$ is in fact a unique $c$, since $h$ is strictly monotone.) But why is it relevant that there's a $b$ such that $f^*(b)=g^*(b)$? That's what I don't understand. (Or maybe there's a lot more that I don't understand.)
What follows are details that I think don't matter, so you can probably ignore the rest of this question.
The two functions $f$ and $g$ are different parameterizations of $$k(x) = \frac{d_1}{d_1+d_2(1-x)+d_3(1-x)^2+(1-d_1-d_2-d_3)(1-x)^3}$$ (e.g. [1] $d_1=0.05$, $d_2=0.30$, $d_3=0.10$; [2] $d_1=0.05$, $d_2=0.20$, $d_3=0.70$). $k(r)$ is in fact a formula for a conditional probability.
Also, in the original case, before composing $f$ and $g$ with $h$ or $h^*$, the values in $(0,1)$ that are arguments to $f$ and $g$ are values of $1-q/p$, where $q/p$ is a likelihood ratio, with $q=$ the probability of a false positive, and $p=$ the probability of a true positive. $h$ and $h^*$ are viewed as transformations of this function of $q$ and $p$. $d_1$, $d_2$, and $d_3$ are functions of the priors. I'll leave out additional detail unless it's needed.
[In case anyone's interested, the paragraph I'm trying to understand is the first full paragraph on page 22; it can be found in the Amazon "search inside" page for the book by searching on "weak bayesian coherentism". Page 20 gives the above formula for k(x), using different notation that will be easy to match up with mine. (I tried to simplify notation for my question.) $q$ and $p$ are defined on pp. 14-15. I can give those definitions explicitly if it's helpful.]