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I am very interested in functions $\gamma:\mathbb R\to\mathbb R$ with the following property: $$\gamma^2(x)=x$$ One form of a function satisfying this is $$f(x)=\frac{a-x}{1+bx}$$ Which has the property $f^2(x)=x$. Infinitely many more functions with this property can be obtained by finding some other injective function $g$, its inverse $g^{-1}$, and then composing $g, g^{-1},$ and $f$ as follows: $$g^{-1}\circ f\circ g$$ However, I am not very interested in involutory functions of this form, since they seem to all be ripoffs of the general form that I already stated.

In fact, it seems that all involutory functions can be put in the form $$g^{-1}\circ f\circ g$$ for some $g$, and for some $a,b$. I can't find any counterexamples, but I don't know how to prove it either. It seems to me that the best way to approach this would be to set up some kind of differential equation like $$(f'\circ f)(x)=\frac{1}{f'(x)}$$ But I have absolutely no idea how I might show that any involutory function can be put in the aforementioned form.

Any ideas?

NOTE: I'm sure there are some elaborate piecewise-defined answers that can destroy my conjecture. However, I can't expect people to know what I mean when I ask to prove this for all "reasonable" functions - so I will establish some stricter restrictions on $\gamma$. The function must be expressible using some finite composition of these functions and their inverses: $$\phi_1(x,a)=x+a$$ $$\phi_2(x,a)=ax$$ $$\phi_3(x,a)=x^a$$ $$\phi_4(x,a)=a^x$$ For example, $x^2+x+1$ can be expressed as $$\phi_1(\phi_3(x,2),\phi_1(x,1))$$

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  • $\begingroup$ what is the role played by $n$? $\endgroup$ – Francesco Polizzi Aug 12 '17 at 16:45
  • $\begingroup$ Whoops, sorry. I had $\gamma^n$ but decided to change it to $2$. I'll fix it! $\endgroup$ – Frpzzd Aug 12 '17 at 16:46
  • $\begingroup$ The answer will very likely depend on your setting. Just sets and functions? Spaces and continuous maps? Groups and homomorphisms? $\endgroup$ – Randall Aug 12 '17 at 16:46
  • $\begingroup$ @Randall I just changed my question to specify that $\gamma$ maps reals to reals. $\endgroup$ – Frpzzd Aug 12 '17 at 16:48
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    $\begingroup$ @Nilknarf If you don't require continuity any involutive bijection can do e.g. the transposition $t_{a,b}$ which exchanges only $a\not=b$. $\endgroup$ – Duchamp Gérard H. E. Aug 12 '17 at 16:51
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There is a general approach using power series. Suppose that $g(x) = b_1x+b_2x^2+b_3x^3+\dots\quad$ and we require that $x+f(x)=g(-xf(x))\;$ for some power series $f(x)$. The connection with $g(x)$ implies $f(f(x))=x.\quad$ Solving for the coefficients of $f(x)$ term by term gives the expansion $$f(x)=-x+b_1x^2-b_1^2x^3+(b_1^3+b_2)x^4-(b_1^4+3b_1b_2)x^5+\dots$$ which is the general form of involution with fixed point $0$ unlike your $f(x)=(a-x)/(1+bx)$ where $f(0)=a$ and $f(a)=0$ with $a\neq 0$. However, If $a=0$ then $$f(x)=-x/(1+bx)=-x+bx^2-b^2x^3+b^3x^4+\dots$$ which is the case where $g(x)=bx$.

You might ask for involutions $f(x)=x+a_2x^2+\dots\;$ but the only example is $f(x)=x$.

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You want an involution $h$ for which no invertible $g$ satisfies $hg=gf$. In fact if $h$ is the identity function this is equivalent to $g=gf$, which by invertibility implies $f$ is the identity function. This fails for any $a,\,b$.

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