if $a,b\in A$, $a\mid b$ then $f(a)\mid f(b)$ we have $\exists m\in A : f(m)=m$ suppose $n\in \Bbb N$ and $A$ is the set of divisors of $n$ and $f:A\to A$ is function such that:
if $a,b\in A$, $a\mid b$ then $f(a)\mid f(b)$
how to prove : $ \exists m\in A : f(m)=m$
it's seems that it has one solution with Pigeonhole principle.
Thanks in advance
 A: Hint: Consider the sequence $1\mid f(1), f(1)\mid f(f(1)),f(f(1))\mid f(f(f(1))), \ldots$
Further hint: Assume that $1\neq f(1), f(1)\neq f(f(1)),f(f(1))\neq f(f(f(1)))\ldots$ to get a contradiction. 
A: Hints:


*

*The divisibility relation is a partial order, and the lattice corresponding to the set of divisors of $n$ is complete.

*In this setting $\forall a,b\in A.\ a \mid b \implies f(a) \mid f(b)$ means that $f$ is order-preserving.

*Your result is a conclusion from Knaster-Tarski theorem. For example, take some $x \in A$, does this sequence $\langle x,f(x),f(f(x)),f^{(3)}(x),\ldots \rangle$ converge?


Good luck!
A: Here is a more general result that fits the problem.


*

*Let B be a finite set (B $\subset \mathbb{N}$) such that there exists u (his greatest element) and v (his smallest element) $ \in$ B that verifies $\forall$ b $\in$ B , v | b | u

*Let $\textit{f}$ :B $\rightarrow $B such that if $a,b\in B$, $a|b$ then $f(a)|f(b)$

*Let us  prove by strong induction on card(B) that $\textit{f}$ has at least one fixed point

*If card(B) = 1, it's trivial

*Suppose the result true for all k $\in$ {0,1,..,n}
If card(B) = n+1 ,


*

*if f(v)=v then it's done

*else, v | $\textit{f}$(v) implies that v < $\textit{f}$(v)
Let B' = B \ {b $\in B$/ $\textit{f}$(b)=v}
u $\in$ B' (if not $\textit{f}$(u)=v implies $\textit{f}$(v)|v which is absurd 
since v< $\textit{f}$(v))


*

*Then B' is a finite set that verifies previous hypothesises, with card(B') < n+1


By our assumption from below, $\textit{f}$ with B' as inputs set has at least one fixed point m


*

*Since B' $\subset$ B , m is also a fixed point for f with B as inputs set

*Hence the result with card(B) = n+1

*Conclusion: the result holds for any B with the previous hypothesises

*Here, u=n and v= 1
