Direct mapping and inverse mapping: when they can be equal?

Question

I read the following phrase from a textbook

Bijection $\ f : M \rightarrow M$ known as symmetry if $\ f^{-1} = f$.

I am curious, are there any examples of $\ f^{-1} = f$ outside of identity relation - $\ id_X $ ?

Answer 1

Here's a really simple example:

Let $X$ be a finite set with $\vert X \vert$ even; then number the elements of $X$ from $1$ to $2n$, $n \in \Bbb N$:

$X = \{x_1, x_2, \ldots, x_n, x_{n + 1}, x_{n + 2}, \ldots, x_{2n} \}; \tag 1$

define

$f:X \to X \tag 2$

by

$f(x_i) = x_{i + n}, \; 1 \le i \le n, \tag 3$

$f(x_i) = x_{i - n}, \; n + 1 \le i \le 2n; \tag 4$

then since $n \ge 2$,

$f \ne \text{Id}_X, \tag 5$

but

$f(f(x_i)) = f(x_{i + n}) = x_i, \; 1 \le i \le n, \tag 6$

$f(f(x_i)) = f(x_{i -n}) = x_i, \; n + 1 \le i \le 2n, \tag 7$

so we see that

$f^2 = \text{Id}_X; \tag 8$

therefore, for $x \in X$,

$f(f(x)) = x, \tag 9$

whence

$f^{-1}(x) = f^{-1}(f(f(x)) = f(x). \tag{10}$

Answer 2

For a set $M$, a function $f:M\to M$ such that $f\circ f=\text{id}_M$ is usually known as an involution on $M$. There exists a one-to-one correspondence between the set $\mathcal{I}$ of all involutions on $M$ and the set $\mathcal{P}$ of partitions of $M$ into subsets of size $1$ or $2$. Such a one-to-one correspondence is given by $\phi:\mathcal{P}\to \mathcal{I}$ sending $P\in\mathcal{P}$ to $$f_P(x)=\begin{cases}x&\text{if }\{x\}\in P\,,\\y&\text{if }\{x,y\}\in P\text{ with }x\neq y\,.\end{cases}$$ The inverse of $\phi$ is $\psi:\mathcal{I}\to \mathcal{P}$ sending $f\in\mathcal{I}$ to $$P_f:=\Big\{\{x\}\,\Big|\,x\in M\text{ and }f(x)=x\Big\}\cup\Big\{\big\{x,f(x)\big\}\,\Big|\,x\in M\text{ and }f(x)\neq x\Big\}\,.$$ If $M$ is a finite set with $m$ elements and $a_m$ denotes the number of involutions on $M$, then $a_0=1$, $a_1=1$, and $$a_m=a_{m-1}+(m-1)a_{m-2}\text{ for }m=2,3,4,\ldots\,.$$ If $M$ is an infinite set, then the cardinality of $\mathcal{I}$ is $2^{|M|}$.