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Do there exist $g,h: \mathbb{N} \rightarrow \mathbb{N}, y \in \mathbb{N}$, such that the following system holds? \begin{align*} (g \circ h \circ h )(y)= 2y \\ (h \circ g \circ h )(y)= 4y \\ (h \circ h \circ g )(y)= 6y \end{align*} I suspect the answer is no, because plugging $h$ into the second and third equation yields $h(2y)=4h(y)$ and $h(4y)=6h(y)$ respectively, and functional equations of the form $f(ax)=bf(x)$ only have the solution $f(x)=cx^{\log_{a}{b}}$, which cannot hold simultaneously for both equations.

Can someone check if this is accurate?

The below answer works for this case, but would it still work if the constants were different (such as $3y,5y,7y$)?

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    $\begingroup$ Functional equations of the form $f(ax)=bf(x)$ have many more solutions if $f$ is only defined on a subset of the rationals. Requiring $f$ to be a continuous function on the reals forces $f(x)=cx^d$ for some constants $c$ and $d$. $\endgroup$ Commented Oct 20 at 13:49

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Suppose $g,h:\ \Bbb{N}\ \longrightarrow\ \Bbb{N}$ are such that \begin{align*} (g \circ h \circ h )(y)= 2y \\ (h \circ g \circ h )(y)= 4y \\ (h \circ h \circ g )(y)= 6y \end{align*} for all $y\in\Bbb{N}$. Then also $h(2y)=4h(y)$ for all $y\in\Bbb{N}$ because $$(h\circ g\circ h\circ h)(y)=(h\circ g\circ h)(h(y)=4h(y),$$ $$(h\circ g\circ h\circ h)(y)=h((g\circ h\circ h)(y))=h(2y),$$ and similarly $h(4y)=6h(y)$ for all $y\in\Bbb{N}$ because $$(h\circ h\circ g\circ h)(y)=(h\circ h\circ g)(h(y)=6h(y),$$ $$(h\circ h\circ g\circ h)(y)=h((h\circ g\circ h)(y))=h(4y).$$ But then $$6h(y)=h(4y)=h(2\cdot2y)=4h(2y)=16h(y),$$ and so $h(y)=0$ for all $y\in\Bbb{N}$, which clearly contradicts all of the required identities.

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  • $\begingroup$ does this argument only work because of the specific choice of constants? that is, if it were $3y,5y,7y$ instead, would this still work? $\endgroup$ Commented Oct 20 at 13:59
  • $\begingroup$ If you cherish the thrilling sensation of doing math with your own hands, read no further, just repeat the argument and plug the new numbers. Otherwise have this: yes, the argument relies heavily on the peculiar identity $4=2\cdot2$ and won't work for an arbitrary combination. $\endgroup$ Commented Oct 21 at 11:47

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