jmc

Let $P(x)=a{x}^4+b{x}^3+c{x}^2+dx+e.$ Then $P(x^2)=a{x}^8+b{x}^6+c{x}^4+d{x}^2+e$ and $$\begin{array}{rl}P(x)P(-x)& =(a{x}^4+b{x}^3+c{x}^2+dx+e)(a{x}^4-b{x}^3+c{x}^2-dx+e)\\ & =(a{x}^4+c{x}^2+e{)}^2-(b{x}^3+dx{)}^2\\ & =(a^2{x}^8+2ac{x}^6+(2ae+c^2)x^4+2ce{x}^2+e^2)-(b^2{x}^6+2bd{x}^4+d^2{x}^2)\\ & =a^2{x}^8+(2ac-b^2)x^6+(2ae-2bd+c^2)x^4+(2ce-d^2)x^2+e^2.\end{array}$$Comparing coefficients, we get $$\begin{array}{rl}{a}^2& =a,\\ 2ac-b^2& =b,\\ 2ae-2bd+c^2& =c,\\ 2ce-d^2& =d,\\ e^2& =e.\end{array}$$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 $$\begin{array}{rl}2c-b^2& =b,\\ -2bd+c^2& =c,\\ -d^2& =d.\end{array}$$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 $$\begin{array}{rl}2c-b^2& =b,\\ 2b+c^2& =c.\end{array}$$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 $$\begin{array}{rl}2c-b^2& =b,\\ 2-2bd+c^2& =c,\\ 2c-d^2& =d.\end{array}$$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 $$\begin{array}{rl}2c-b^2& =b,\\ 2-2b^2+c^2& =c.\end{array}$$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 $$\begin{array}{rl}{x}^4& =x^4,\\ x^4-x^3& =x^3(x-1),\\ x^4+x^3+x^2& =x^2(x^2+x+1),\\ x^4-2x^3+x^2& =x^2(x-1{)}^2,\\ x^4-x& =x(x-1)(x^2+x+1),\\ x^4-3x^3+3x^2-x& =x(x-1{)}^3,\\ x^4-4x^2+6x^2-4x+1& =(x-1{)}^4,\\ x^4-x^3-x+1& =(x-1{)}^2(x^2+x+1),\\ x^4+x^3+x^2+x+1& =x^4+x^3+x^2+x+1,\\ x^4+2x^3+3x^2+2x+1& =(x^2+x+1{)}^2.\end{array}$$