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 $.
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.
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.