$f$ be a bijection on a finite set $X$ , moving at most finitely many points , then is $f^n$ the identity map for some positive integer $n$? Let $X$ be an infinite set and $f :X \to X $ be a bijection such that there exist a finite subset $A \subseteq X$ such that $f(x)=x , \forall x \in X \setminus A$ ; then does there exist $ n \in \mathbb N$ such that $f^n$ is the identity map ? 
 A: Let $m$ be the size of $A$. Then $f \,|\, A$ is (equivalent to) a permutation in $S_m$. Let $n=m!$. By Lagrange's theorem, $(f \,|\, A)^{n}=id_A$. Therefore, $f^{n}=id_X$.
If you don't want to use Lagrange's theorem from group theory, argue as follows.
Since there are only a finite number of bijections of $A$, the powers of  $f \,|\, A$ must repeat. Since $f$ is a bijection this implies that $(f \,|\, A)^{n}=id_A$ for some $n$.
A: Yes, let $A=\{a_1, a_2, \ldots, a_n\}$ and $f:X\rightarrow X$ be a bijection that fixes $X\setminus A$. Then, for each $i=1,2,\ldots, n$, we can write represent $f$ as disjoint cycles. For example: Let $\mathbb{N}$ and suppose $A=\{1,2,\ldots, 8\}$. Say $f(1)=2, f(2)=3, f(3)=1, f(4)= 6, f(5)=7, f(6)=5, f(7)=8, f(8)=4$. Then we may represent $f=(1,2,3)(4,6,5,7,8)$. We call $(1,2,3)$ and $(4,6,5,7,8)$ cycles. In general, we can write any such $f$ as a 'product of disjoint cycles'. If we define the length of the cycle $(1,2,3)$ as $3$ and similar to any other cycle, then the following holds:
Suppose $f$ is represented as a product of disjoint cycles $\mu_1 \mu_2 \mu_3\ldots \mu_r$ with  $\mu_j$ a cycle of length $n_j$ for each $j =1,2,\ldots, r$. Letting $n=lcm\{n_j\ |\ j=1,2,\ldots, r\}$, $n$ is the smallest positive integer for which $f^n=id$.
You can check any introductory text on Group theory or abstract algebra for a formal discussion of cycles and permutations of a finite set.
A: Let $S:=\{x \in X : f(x) \ne x\}$ ; then $S \subseteq A$ , so $S$ is a finite set . Now , since $f$ is an injection , so $x \in S \implies f(x) \ne x \implies f(f(x)) \ne f(x) \implies f(x) \in S$ ; thus $f: S \to S$ is a well-defined restriction of an injective map , so this restriction is injective , and since $S$ is finite , so this restriction is bijective also ; thus there is $n \in \mathbb N$ such that $f^n(x)=x , \forall x \in S$ , then $f^n(x)=x , \forall x \in X$ 
