The trivial case is H(x)=0.

Let H be of degree n. Then H' and H" are of degrees n-1 and n-2 respectively. Then since deg H = deg (H'*H"), n = (n-1)+(n-2), or n=3.

Let H(x)=ax³+bx²+cx+d and differentiate twice. Multiply H' and H" together to get H'*H"=18a²x³+18abx²+(4b²+6ac)x+2bc.

Substitute 2x into H to get H(2x)=8ax³+4bx²+2cx+d.

Comparing cubic coefficients, it must be the case that a=4/9.

Comparing quadratic coefficients, it must be the case that a=2/9 (which can not be), or that b=0.

Comparimg linear coefficients, it must be the case that, since b=0, then 2c=6ac, hence a=1/3 (which can not be), or that c=0.

Finally, if d=2bc and b=c=0, then d=0.

Thus the only nontrivial polynomial that satisfies the equation is H(x)=4x³/9.