If $f$ is strictly decreasing, then we first have either:
1) $x < f(x)$
2) $x > f(x)$
1) If $x < f(x)$, we know $f^2$ is strictly increasing, so $x < f(x) < f^2$. But $f^3$ is decreasing, so $f^3 < f^2$ but we don't know its relation to $x$ and $f(x)$. $f^4$ is increasing again, so $f^4 > f^3$ but if $f^3 < f(x)$, then it's possible that $f^4 < f^2$, but it's also possible that if $f(x) < f^3 < f^2$, then it's possible that $f^4 \geq f^2$, so it seems impossible to tell anything about the orbits. So it seems like we can't tell if this $f$ has any fixed points, points of prime order, or if there's an infinite orbit. How can we show that $f$ can have only fixed points, points of prime order $2$, and/or points whose orbits are infinite?