It should at least be mildly amusing that such functions exist. For example the functions $1/x$ and $1 - x$ are both self-inverse, $f^{\circ2}(x)=x$. Now let's do a mix and match on these two functions to get $g(x)= 1-1/x$ and $h(x) = 1/(1-x)$. Neither of these functions is self-inverse, but $g^{\circ3}(x) = h^{\circ3}(x) = x$. Further, $g^{\circ2} = h$ and $h^{\circ2} = g$.
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See, this question for example. – lulu Apr 25 '23 at 20:01
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1These are sometimes referred to as "cyclical functions (of order 3)." See here for such a usage. Certainly the concept is well studied in the context of algebra and permutations, but I personally don't usually encounter much interest in it outside of those settings. (Of course, that may say more about me than about the problem). – JMoravitz Apr 25 '23 at 20:03
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3Several very relevant Wikipedia links. – J.G. Apr 25 '23 at 20:03
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A very recent question here. – Jean Marie Apr 25 '23 at 20:36