The simplest polynomial in x is a linear one, such as 3x - 5, and its derivative is 3. Finding the derivative of a polynomial is simple: multiply each term by its power in x, and reduce the power by 1. Since the power in x of every term is being reduced by 1, the derivative of the function will never be equal to the function... for a finite-order function. If there is no highest order term, i.e. the order is infinite, then it is possible.
For p(x) = p'(x) to be true, each term (a_n)x^n in p(x) (where a_n means 'a subscript n' and a_n is the coefficient of x^n) must be equal to the derivative of the next term (a_(n+1))x^(n+1) in p(x), which is (n+1)(a_(n+1))x^n. So:
(a_n)x^n = (n+1)(a_(n+1))x^n => a_n = (n+1)(a_(n+1)) => a_(n+1) = (a_n)/(n+1)
So the coefficients of x^n follow a sequence, such that the next coefficient (that of x^(n+1)) is equal to the previous coefficient divided by n+1. If we let a_1 = 1:
a_1 = 1, a_2 = 1/2, a_3 = 1/6, a_4 = 1/24 => a_n = 1/n!
So the polynomial p(x) = 1 + x + (x^2)/2 + (x^3)/6 + ... + (x^n)/n!
This turns out to be the Maclaurin series expansion (a polynomial approximation) of e^x, the simplest function such that f(x) = f'(x).