Let $f: [0,1] \to [0, 1]$ be a continuous function.
Let $Fix(f^2)=\{x \in [0,1]: f^2(x)=x\}$ (where $f^2=f \circ f $),$\hspace{0.1cm}$ $Fix(f)=\{x \in [0,1]: f(x)=x\}$ and $Per_2(f)=\{x \in [0,1]: mín\{n \in \mathbb{N}: f^n(x)=x\}=2\}$.
If $Fix(f^2)$ is connected, then $Fix(f)$ is also connected.
Here is what I've attempted so far (let it be noted that I'm aiming for a contradiction):
It is obvious that $Fix(f^2)=Fix(f) \cup Per_2(f)$ and this union is disjoint. I've already proven that $Fix(f)$ is closed, if $Per_2(f)$ were also closed then $Fix(f^2)$ is the union of two disjoint, closed sets; therefore $Fix(f^2)$ is not connected, which is the contradiction I'm looking for. However, $Per_2(f)$ need not be closed: $f: [0,1] \to [0, 1]$, $\hspace{0.1cm}$ $f(x)=\sqrt{1-x^2}$ is a continuous function where $Fix(f^2)$ is connected, $Per_2(f)$ is not closed and $Fix(f)$ is connected.
From here on out everything else I've tried doesn't seem to yield any concrete results. I've noticed that I've yet to use that $f$ is continuous, so that might be why I'm unable to get to the contradiction I need in order to finish the proof.
Any tip to help me to get on the right track would be much appreciated.
Let $\Phi_k = \{ x | f^k(x) = x \}$. Then $\Phi_1 \subset \Phi_2$ and both are closed by continuity of $f$.
Suppose $\Phi_2$ is connected, then it is a closed interval. Suppose $\Phi_1$ is not an interval. In particular, there are points $x_1 < x^*< x_2$ such that $f(x_1) = x_1, f(x^*) \neq x^*, f(x_2) = x_2$. We can assume that $x_1$ is the largest value less than $x^*$ such that $f(x_1)= x_1$ and similarly for $x_2$.
Assume for the moment that $f(x^*) < x^*$, and so $f(x) < x$ for all $x \in (x_1,x_2)$. Choose $z \in [x_1,x_2]$ such that $f([z,x_2]) = [x_1,x_2]$.
Pick some $x \in (z,x_2)$. Then $f(x) < x$, and since $f(x) \in [x_1,x_2]$, we have $f(f(x)) < f(x)$ which contradicts $\Phi_2$ being an interval.
If $f(x^*) >x^*$, a similar argument results in the same contradiction. Hence $\Phi_1$ is an interval.