Accept complex number.

f(x)=a(x-2)(x-b) a is not 0.

case 1: f(5) = 2 => b=(15a-2)/3a => f(0)=2ab=10a - 4/3.

case 2: f(5) = b => b=15a/(3a+1) => f(0)=2ab=30a/(3a+1).

Discussion:

let f(f(x)) = a^3(x-5)(x-p)(x-q)(x-r).

case 1: we have f(x) - 2 = a(x-5)(x-p) and f(x) - b = a(x-q)(x-r)

case 2: we have f(x) - b= a(x-5)(x-p) and f(x) - 2 = a(x-q)(x-r).

If a is real, in either case, b should be real, and we already know the unique answer.

If a is unreal, in either case, b is unreal too. Easy to see p, q and r are all unreal.

That is, a could be any unreal and f(0) could be any unreal.