Let $f(x) = \frac{x + 6}{x}.$  The sequence $(f_n)$ of functions is defined by $f_1 = f$ and
\[f_n = f \circ f_{n - 1}\]for all $n \ge 2.$  For example,
\[f_2(x) = f(f(x)) = \frac{\frac{x + 6}{x} + 6}{\frac{x + 6}{x}} = \frac{7x + 6}{x + 6}\]and
\[f_3(x) = f(f_2(x)) = \frac{\frac{7x + 6}{x + 6} + 6}{\frac{7x + 6}{x + 6}} = \frac{13x + 42}{7x + 6}.\]Let $S$ be the set of all real numbers $x$ such that
\[f_n(x) = x\]for some positive integer $n.$  Find the number of elements in $S.$
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.