Joel's answer made me think a bit and I believe I found an interesting solution for $f(x)$ :
$$ f(x) = \begin{cases} ix & \text{if } \mathrm{Im}(x) = 0, x\neq 0 \\ \cos(ix) & \text{if } \mathrm{Re}(x) = 0,x \neq 0 \\ 2\pi i & \text{if } x = 0 \end{cases}$$
It is of course a bit of a trick (reminds me of Wick Rotation), but I it works for all $x \in \mathbb R$, because
$$f(f(x)) = \cos(i(ix))=\cos(-x) = \cos(x)$$
Update: Added the case $x=0$. For this we have
$$f(f(0)) = \cos(i(2\pi i))=\cos(-2\pi) = \cos(0)$$
