# Are all commuting functions "the same"?

Let $f,g,h$ be "nice" functions and $h(H(x))=H(h(x))=x$. Let $F(x)=H(f(h(x)))$ and $G(x)=H(g(h(x)))$. Obviously $f(g(x))=g(f(x)) \Leftrightarrow F(G(x))=G(F(x))$.
Example: $f(x)=ax, g(x)=bx$ and $F(x)=a+x,G(x)=b+x$ are related by a log transformation.
Question: Can all commuting functions $f,g$ be transformed to $F(x)=a+x,G(x)=b+x$ with some $h$? You may restrict "nice" as you like (monotonous? differentiable? complex domain?) if a proof demands it. (It's probably not hard to come up with an un-nice counterexample.)

• I wouldn't say this answers your question, but a positive result along these lines is the theorem that a family of commuting diagonalizable matrices is simultaneously diagonalizable: if such matrices commute then they are simultaneously similar to some matrices which obviously commute. Aug 13, 2013 at 16:46

Let me rephrase your question: is every pair of commuting functions simultaneously conjugate to a pair shifts? (Your conditions, literally understood imply that your $h$ is a bijective map.) The answer is evidently no.
In your example, $ax$ and $bx$ are restricted to the positive ray. If you consider the "same" functions on the whole real line, they are already not conjugate to shifts. Because they have a fixed point while the shifts do not.
One can replace conjugacy by semiconjugacy, requiring in your notation $h(F)=f(h)$, and $h(G)=g(h)$ (without the requirement that $h$ is invertible) but even in this case, not every pair of commuting functions is semiconjugate to an affine pair.