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I can think of several examples of functions such that twice application of the function is equivalent to no application of it.

  • Additive inverse
  • Multiplicative inverse
  • Fourier transform
  • Complex conjugation
  • Any group built up from $\mathbb{Z}_2$, applying (one of) the $\mathbb{Z}_2$ parts' operation.

"Idempotent" came to mind, but that's wrong. It means $f(f(x)) = f(x)$, not $f(f(x))=x$.

What is the word for this "flip-flop" property?

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    $\begingroup$ Involution $\endgroup$ – Brandon Carter Nov 26 '11 at 19:12
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    $\begingroup$ also known as '$f$ is its own inverse' $\endgroup$ – Chris Taylor Nov 26 '11 at 19:13
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    $\begingroup$ @BrandonCarter Why did you comment instead of answering? I'm not sure whether to "award" the checkmark to you or Leandro. $\endgroup$ – isomorphismes Nov 26 '11 at 19:41
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    $\begingroup$ Fourier transform doesn't have that property ;) You get an extra reflection.... $\endgroup$ – N. S. Nov 26 '11 at 20:21
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    $\begingroup$ possible duplicate of Functions that are their own inversion. $\endgroup$ – Najib Idrissi Aug 31 '15 at 8:39
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I think this is called an "Involution".

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  • $\begingroup$ this is the accepted answer. therefore, please elaborate more and/or cite your sources. $\endgroup$ – chharvey Dec 31 '16 at 6:11
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Involution is the most common name. They are so fundamental that an entire book has been written on them, the Book of Involutions. I often emphasize their essential role both here and various other places. One should always strive to bring to the fore the innate symmetries in problems, and involutions are one of the simplest examples.

Note: you could have found the answer simply by Googling "self inverse function". The first match is the "self inverse" section of the Wikipedia page on inverse functions, which states "such a function is called an involution".

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    $\begingroup$ Self inverse. I didn't think of that. I googled for something approximately like the title. (the word self inverse didn't come readily to mind because I was thinking $f^2=f^0$ rather than $f^1 = f^{-1}$) $\endgroup$ – isomorphismes Nov 26 '11 at 21:02
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    $\begingroup$ @Lao It is well-worth the effort to learn how to compose effective searches, by means of which you can truly stand on the shoulder of giants - exposing many beautiful mathematical vistas. $\endgroup$ – Bill Dubuque Nov 26 '11 at 21:09

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