jmc

First, we solve the equation $f(x)=x.$ This becomes $$\frac{x+6}{x}=x,$$so $x+6=x^2,$ or $x^2-x-6=(x-3)(x+2)=0.$ Thus, the solutions are $x=3$ and $x=-2.$

Since $f(x)=x$ for $x=3$ and $x=-2,$ $f_n(x)=x$ for $x=3$ and $x=-2,$ for any positive integer $n.$ Furthermore, it is clear that the function $f_n(x)$ will always be of the form $$f_n(x)=\frac{ax+b}{cx+d},$$for some constants $a,$ $b,$ $c,$ and $d.$ The equation $f_n(x)=x$ then becomes $$\frac{ax+b}{cx+d}=x,$$or $ax+b=x(cx+d).$ This equation is quadratic, and we know it has roots 3 and $-2,$ so there cannot be any more solutions to the equation $f_n(x)=x.$

Therefore, $S=\{3,-2\},$ which contains $\boxed{2}$ elements.