jmc

Let $n$ be the degree of $p(x).$ Then the degree of $p(p(x))$ is $n^2,$ and the degree of $xp(x)$ is $n+1.$

If $n\ge 2,$ then the degree of $xp(x)+x^2$ is $n+1,$ which is strictly less than $n^2.$ Also, $p(x)$ clearly cannot be a constant polynomial, so the degree of $p(x)$ is $n=1.$

Let $p(x)=ax+b.$ Then $$p(p(x))=p(ax+b)=a(ax+b)+b=a^{2}x+ab+b,$$and $$xp(x)+x^2=x(ax+b)+x^2=(a+1)x^2+bx.$$Equating coefficients, we get $a+1=0,$ $a^2=b,$ and $ab+b=0.$ Then $a=-1$ and $b=1,$ so $p(x)=\boxed{-x+1}.$