I made the following problem a while ago but I can't solve it (also I don't think it's extremely hard ) :
Let $f$ be a non-constant completely multiplicative function over $\mathbb{Z}$ . Assume that there is some non-constant polynomial $P \in Z[X]$ such that $P \circ f $ is also a completely multiplicative function . Then prove that $P$ is a monomial (something like $P(x)=ax^m$ ) .
If you want a challenge then replace completely multiplicative with just multiplicative . Also you can try with $f \circ P$ instead of $P \circ f$ (though I think this makes the problem simpler )
This is easily one of my favorites problems (partly because I came with it myself ) but I couldn't solve it .
Thanks for all the help .
EDIT In response to user270395's answer I'll add the additional condition :
$$f(n) \neq 0$$ for every $n \neq 0$ .
I hope this suffices .
EDIT 2 I feel embarrassed that my previous two problems had counter-examples but I think the following new condition is strong enough to get past the "trivial" counter-examples (thanks to EricWong for his thoughts) :
$f(n)$ takes an infinite number of different values and all are non-zero (except $f(0)=0$) .
Its not true. Let $f(\pm 1)=1$ and $f(n)=0$ for all other $n$. Let $P(x)=x^2-x+1$ then $P(f(x))=1$ for all $x$.