jmc

We have that $$\begin{array}{rl}f(x)f(-x)& =(a{x}^2+bx+c)(a{x}^2-bx+c)\\ & =(a{x}^2+c{)}^2-(bx{)}^2\\ & =a^2{x}^4+2ac{x}^2+c^2-b^2{x}^2.\end{array}$$We want this to equal $f(x^2)=a{x}^4+b{x}^2+c.$ Comparing coefficients, we get $$\begin{array}{rl}{a}^2& =a,\\ 2ac-b^2& =b,\\ c^2& =c.\end{array}$$Thus, $a=0$ or $a=1,$ and $c=0$ or $c=1.$ We divide into cases accordingly.

If $a=0$ or $c=0,$ then $ac=0,$ so $$b^2+b=b(b+1)=0,$$which means $b=0$ or $b=-1.$

The only other case is where $a=1$ and $c=1.$ Then $$b^2+b-2=0,$$which factors as $(b-1)(b+2)=0.$ Hence, $b=1$ or $b=-2.$

Therefore, there are $\boxed{8}$ such functions $f(x)$: $$0,1,-x,1-x,x^2,x^2-x,x^2+x+1,x^2-2x+1.$$