Let $ P(x) = ax^4 + bx^3 + cx^2 + dx + e $. Then $ P(x^2) = ax^8 + bx^6 + cx^4 + dx^2 + e $ and \begin{align*} P(x) P(-x) &= (ax^4 + bx^3 + cx^2 + dx + e)(ax^4 - bx^3 + cx^2 - dx + e) \\ &= (ax^4 + cx^2 + e)^2 - (bx^3 + dx)^2 \\ &= (a^2 x^8 + 2acx^6 + (2ae + c^2) x^4 + 2cex^2 + e^2) - (b^2 x^6 + 2bdx^4 + d^2 x^2) \\ &= a^2 x^8 + (2ac - b^2) x^6 + (2ae - 2bd + c^2) x^4 + (2ce - d^2) x^2 + e^2.\end{align*}Comparing coefficients, we get \begin{align*} a^2 &= a, \\ 2ac - b^2 &= b, \\ 2ae - 2bd + c^2 &= c, \\ 2ce - d^2 &= d, \\ e^2 &= e.\end{align*}From $ a^2 = a, $ $ a = 0 $ or $ a = 1 $. But $ P(x) $ has degree 4, which means that the coefficient of $ x^4 $ cannot be 0, so $ a = 1 $. From $ e^2 = e, $ $ e = 0 $ or $ e = 1 $. Case 1: $ e = 0 $. The equations become \begin{align*} 2c - b^2 &= b, \\ -2bd + c^2 &= c, \\ -d^2 &= d.\end{align*}From $ -d^2 = d, $ $ d = 0 $ or $ d = -1 $. If $ d = 0, $ then $ c^2 = c, $ so $ c = 0 $ or $ c = 1 $. If $ c = 0, $ then $ -b^2 = b, $ so $ b = 0 $ or $ b = -1 $. If $ c = 1, $ then $ 2 - b^2 = b, $ so $ b^2 + b - 2 = (b - 1)(b + 2) = 0, $ which means $ b = 1 $ or $ b = -2 $. If $ d = -1, $ then \begin{align*} 2c - b^2 &= b, \\ 2b + c^2 &= c.\end{align*}Adding these equations, we get $ 2b + 2c - b^2 + c^2 = b + c, $ so \[b + c - b^2 + c^2 = (b + c) + (b + c)(-b + c) = (b + c)(1 - b + c) = 0.\]Hence, $ b + c = 0 $ or $ 1 - b + c = 0 $. If $ b + c = 0, $ then $ c = -b $. Substituting into $ 2c - b^2 = b, $ we get $ -2b - b^2 = b, $ so $ b^2 + 3b = b(b + 3) = 0 $. Hence, $ b = 0 $ (and $ c = 0 $) or $ b = -3 $ (and $ c = 3 $). If $ 1 - b + c = 0, $ then $ c = b - 1 $. Substituting into $ 2c - b^2 = b, $ we get $ 2b - 2 - b^2 = b, $ so $ b^2 - b + 2 = 0 $. This quadratic has no real roots. Case 2: $ e = 1 $. The equations become \begin{align*} 2c - b^2 &= b, \\ 2 - 2bd + c^2 &= c, \\ 2c - d^2 &= d.\end{align*}We have that $ 2c = b^2 + b = d^2 + d, $ so \[b^2 - d^2 + b - d = (b - d)(b + d) + (b - d) = (b - d)(b + d + 1) = 0.\]Hence, $ b = d $ or $ b + d + 1 = 0 $. If $ b + d + 1 = 0, $ then $ d = -b - 1 $. Substituting into $ 2 - 2bd + c^2 = c, $ we get \[2 - 2b(-b - 1) + c^2 = c,\]so $ 2b^2 + 2b + c^2 - c + 2 = 0 $. Completing the square in $ b $ and $ c, $ we get \[2 \left( b + \frac{1}{2} \right)^2 + \left( c - \frac{1}{2} \right)^2 + \frac{5}{4} = 0,\]so there are no real solutions where $ b + d + 1 = 0 $. If $ b = d, $ then the equations become \begin{align*} 2c - b^2 &= b, \\ 2 - 2b^2 + c^2 &= c.\end{align*}From the first equation, $ c = \frac{b^2 + b}{2} $. Substituting into the second equation, we get \[2 - 2b^2 + \left( \frac{b^2 + b}{2} \right)^2 = \frac{b^2 + b}{2}.\]This simplifies to $ b^4 + 2b^3 - 9b^2 - 2b + 8 = 0, $ which factors as $ (b + 4)(b + 1)(b - 1)(b - 2) = 0 $. Hence, the possible values of $ b $ are $ -4 $, $ -1, $ 1, and 2, with corresponding values of $ c $ of 6, 0, 1, and 3, respectively. Thus, there are $ \boxed{10} $ polynomials $ P(x), $ namely \begin{align*} x^4 &= x^4, \\ x^4 - x^3 &= x^3(x - 1), \\ x^4 + x^3 + x^2 &= x^2 (x^2 + x + 1), \\ x^4 - 2x^3 + x^2 &= x^2 (x - 1)^2, \\ x^4 - x &= x(x - 1)(x^2 + x + 1), \\ x^4 - 3x^3 + 3x^2 - x &= x(x - 1)^3, \\ x^4 - 4x^2 + 6x^2 - 4x + 1 &= (x - 1)^4, \\ x^4 - x^3 - x + 1 &= (x - 1)^2 (x^2 + x + 1), \\ x^4 + x^3 + x^2 + x + 1 &= x^4 + x^3 + x^2 + x + 1, \\ x^4 + 2x^3 + 3x^2 + 2x + 1 &= (x^2 + x + 1)^2.\end{align*}