Problem in Divisibility and Functions Suppose that $ n $ is an integer, $ A $ is the set of $ n $'s divisors and $ f :A\rightarrow A $ be a function with this property: if $ a $ and $ b $ belong to $ A $ and $ a\mid b $, then $ f (a)\mid f (b) $. Prove that there is an integer $m$ in $ A $ that $f (m)=m $.
 A: If $n$ is a prime number than it is clear, because either $f$ is constant or the identity.
If $n$ is not a prime number:Assume that f does not have a fixed point and  let $f(1) = m$. Then we can define $f'$ for $n/m$ by $f'(k) = f(k)/m$ for all divisors $k$ of $n/m$. Since $f$ has no fixed point, also $f'$ has no fixed point. Inductively we will arrive at a prime and get a contradiction. 
A: We have $1 \mid f(1) \mid f^2(1) \mid f^3(1) \mid \cdots$
Since $A$ is finite, this sequence must repeat: $f^{j+p}(1)=f^j(1)$, for some $j \ge 0$ and $p \ge 1$.
Since $f$ is increasing, we must have $p=1$, if $p$ is the minimal period.
Then $m=f^j(1)$ is a fixed point of $f$: $f(m)=m$.
A: $f$ is clearly increasing. So you can use the fact that an increasing function from $A$ to $A$ must have a fixpoint.
Let $ B = \{ x \in A \mid f(x) > x \}$. If $B$ is empty, then $1 \not\in B$ thus $f(1) \leq 1$ but since $f(1)$ must be in $A$ too, $f(1) \geq 1$ and thus $m=1$ do the job.
Otherwise, you can consider $m = max(B)$, then $f(f(m)) \geq f(m) > m$ because $f$ is increasing and $m \in B$. Now assume $f(f(m)) > f(m)$, then $f(m)$ would be in $B$ with $f(m) > m = max(B)$ : this is a contradiction. Hence, $f(f(m)) \leq f(m)$ and $f(m)$ is the fixpoint you are looking for.
A: I'm assuming that you intended to say that "... there is an integer $m$ ...", as @BrianTung pointed out.
However, the statement then isn't true.  Take $n=1,$ so that $A = \lbrace 1,-1 \rbrace.$ Define $f(x)=-x$ for a counterexample.
I suspect that the problem was intended to be about positive integers.
