Let $X$ be an infinite set. Prove $\exists f : X \rightarrow X$ s.t $\forall x \in X, \forall n > 0, f^n(x)\neq x$ I'm reading "Sets, Models and Proofs" by I. Moerdijk and J. van Oosten to have some basic understanding of set theory. 
There's an exercise I don't see how to work it out:
Exercise 20 Let $X$ be an infinite set. Prove that there is a bijection $f : X \rightarrow X$ with the property that for every $x \in X$ and all $n > 0$, $f^n(x)\neq x$ [Hint: consider $\mathbb{Z} \times X$, or use Zorn directly].
How to prove that?
 A: First prove that for $\Bbb Z$. Which is quite easy.
Then note that if we can show this property for $Y$, then whenever $|X|=|Y|$ we can show this property for $X$ as well. Finally, fix a bijection between $X$ and $X\times\Bbb Z$, and apply the function from the first step on the $\Bbb Z$-coordinates.
A: "fix a bijection between $X$ and $X×\mathbb{Z}$"
$\mathbb{Z}$ has the smallest cardinal number among infinite sets.
For any infinite set $X$, we have the cardinal number of $X\times\mathbb{Z}$ the same as $X$. (Direct result from some theorem I don't remember well.) There exists a bijection $g$ between $X$ and $X×\mathbb{Z}$ since the two infinite set have the same cardinal number.
$∃f:Z→Z s. t. ∀x∈Z,∀n>0,fn(x)≠x $ -- just let $f(x)=x+1$.
Apply $f$ on the second element of the pair $(X,\mathbb{Z})\in X\times\mathbb{Z}$
$g\circ f\circ g^{-1}$ is a bijection for $X$ satisfies the condition
since
$(g\circ f\circ g^{-1})^n=g\circ f^{(n)}\circ g^{-1}$

Addition on 27th April 2014
On the proof that $X$ and $\mathbb{Z}\times X$ have the same cardinal number
Or on the construction of bijection between $X$ and $\mathbb{Z}\times X$
First of all, we construct a bijection from $\mathbb{Z}^+\times X^+$ to $X^+$
$f:\mathbb{Z}^+\times X^+\rightarrow X^+$
For each pair $(m\in\mathbb{Z}^+,n\in  X^+)$
We assume


*

*$X^+$ is always a set of numbers, if it is not a set of numbers, we can always replace it with a set of numbers having the same cardinal number. 

*n as a number is always in decimal representation, where $[n]\geq 0$ is the floor of $n\geq 0$, namely the largest integer no larger than n with $1\geq n-[n]\geq 0$.


$f((m\in\mathbb{Z}^+,n\in  X^+))= \frac{(m+[n])(m+[n]+1)}{2}+n$
Then we construct a bijection from $\mathbb{Z}^+\times X$ to $X$
$g:\mathbb{Z}^+\times X\rightarrow X$
$g((m\in\mathbb{Z}^+,n\in X))=f(m\in\mathbb{Z}^+,n\in X^+)$ if $n\geq 0$
$g((m\in\mathbb{Z}^+,n\in X))=-f(m\in\mathbb{Z}^+,|n|\in X^+)$ if $n\leq 0$
Then we construct a bijection from $\mathbb{Z}\times X$ to $X$
First of all we construct a bijection $h:\mathbb{Z}\rightarrow \mathbb{Z}^+$
$h(x\in \mathbb{Z})=2x$ if $x>0$
$h(x\in \mathbb{Z})=2|x|+1$ if $x\leq 0$
Finally
$g((h(m\in\mathbb{Z}),n\in X))$
is a bijection from $\mathbb{Z}\times X$ to $X$

Clarifications on "replace the infinite set with a set of numbers"
I intended to say "replace the infinite set with a set of real numbers with the same cardinal number"
This is not rigourous statement in mathematics.
Here I replace it with a rigourous one.
If the axiom of choice holds, then a set is infinite if and only if it includes a countable infinite subset.
Set of integer is a countable infinite set. And countable infinite set has the smallest cardinal number among all the infinite set.
Let $X$ be an infinite set, and let $A=X\cup\mathbb{Z}$, then $A$ has the same cardinal number as $X$
This is easy to prove by establishing a bijection between $X$ and $A$.
For any infinite set $X$, let $x\in X$
We have $\{x\}\cup \mathbb{Z}$ is countable infinite and has the same cardinal number as $\mathbb{Z}$. We denote $\{x\}\cup \mathbb{Z}$ as $\mathbb{Z}_x$
$\mathbb{Z}\times X=\bigcup\mathbb{Z}\times x|\forall x\in X$
There exists a bijection between $\mathbb{Z}\times x$ and $\mathbb{Z}_x$ for $\forall x\in X$, since both sets are countable infinite.
There exists a bijection between $\bigcup\mathbb{Z}\times x|\forall x\in X$ and $\bigcup\mathbb{Z}_x|\forall x\in X$
since $\bigcup\mathbb{Z}_x|\forall x\in X=X\cup\mathbb{Z}$
then there exists a bijection between $\mathbb{Z}\times X$ and $X$.
Proved.
A: Following Asaf's hint. Let me have a try.
First, $\exists f:\mathbb{Z}\rightarrow\mathbb{Z}$ s. t. $\forall x\in \mathbb{Z},\forall n>0, f^n(x)\ne x$ -- just let $f(x)=x+1$.
Second, a poset $P: (X, f, \le)$ can be defined as bijection $f:U\rightarrow U$, $U\subset X$, $(\forall n>0)(\forall x\in X)(f^n(x) \ne x)$, and $(U, f)\le(V,g)$ iff $U\subset V$ and $u$ is the restriction of $v$ to $U$. 
$P$ is not empty due to the first step. 
Also, if $\{(U_i, f_i) \mid i\in I \}$ is a chain in $P$, there is a well-defined function $f:\cup_i U_i \rightarrow \cup_i U_i$ is a bijection. So any chain in $P$ has an upper bound.
Now due to Zorn's lemma, such $P$ has a maximal element $(U, f)$. 
Third, if $U \ne X$, it shall be either $X-U$ is infinite or finite.
i) if $X-U$ is finite, a bijection $g:X\rightarrow U$ can be defined, then as $X\xrightarrow{g}U\xrightarrow{f}U\xrightarrow{g^{-1}}X$, $h:= g^{-1}\circ f\circ g : X\rightarrow X$ is a bijection that $h^n(x) \ne x$. we are done.
ii) if $X-U$ is infinite, a bijection $Y:V\rightarrow V$ can be found similiarly on $V\subset X-U$, then $h:=f+g$ could be defined, this conflicts with the second step that $(U,f)$ is the maximal element on $P$.
so we are done.
