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Let $m$ and $n$ be the degrees of $P(x)$ and $Q(x),$ respectively. Then the degree of $P(Q(x))$ is $mn.$ The degree of $P(x)Q(x)$ is $m+n,$ so $$mn=m+n.$$Applying Simon's Favorite Factoring Trick, we get $(m-1)(n-1)=1,$ so $m=n=2.$

Let $P(x)=a{x}^2+bx+c.$ From $P(1)=P(-1)=100,$ $a+b+c=100$ and $a-b+c=100.$ Taking the difference of these equations, we get $2b=0,$ so $b=0.$ Then from the given equation $P(Q(x))=P(x)Q(x),$ $$aQ(x{)}^2+c=(a{x}^2+c)Q(x).$$Then $$c=(a{x}^2+c)Q(x)-aQ(x{)}^2=(a{x}^2+c-aQ(x))Q(x).$$The right-hand side is a multiple of $Q(x),$ so the left-hand side $c$ is also a multiple of $Q(x).$ This is possible only when $c=0.$

Hence, $a=100,$ so $P(x)=100x^2,$ which means $$100Q(x{)}^2=100x^{2}Q(x).$$Cancelling $100Q(x)$ on both sides, we get $Q(x)=\boxed{x^2}.$