Lemma Let $\phi : [0,1] \to [0,1]$ is a continuous function. Let $x_0$ is fixed point of $\phi$ i.e. $\phi(x_0)=x_0$ . Let $x_1 \in [0,1]$ is such that $x_1$ is between $x_0$ and $\phi(x_1)$, i.e $\phi(x_1) < x_1 < x_0$ or $x_0 < x_1 < \phi(x_1)$. Then for each $n \in \mathbb{N}$ there exists $y \in [0,1]$, such that $ M=\{\phi^n(y)\}_{n=1}^{\infty}$ consists of more than $n$ different points. Proof. Let first assume $x_0 <x_1 < \phi(x_1)$ Since $\phi(x_0)=x_0<x_1,\, \phi(x_1)>x_1$ and $\phi$ is continuous, there exists $x_2 \in (x_0, x_1), \, \phi(x_2)=x_1$. If apply this again to $x_0, x_2, x_1=\phi(x_2)$, there will be $x_3, \in (x_0,x_2),\, \phi(x_3)=x_2$. Thus we get $\{x_j\}_{j=1}^{\infty}, \, x_{j+1} < x_j,\, \phi(x_{j+1})=x_j$. Now it is clear that for each $n \in \mathbb{N}$ we can get sufficiently big $j$, so that $\{\phi^n(x_j)\}_{n=1}^{\infty} $ consists of more than $n$ different points. The case $\phi(x_1) < x_1 < x_0$ is similar. There exist at leasts one fixed point of $f$. It follows from a general Brower's result, or in this case can be simply obtained by the continuity of $f_1(x)=f(x)-x,\, f_1(0) \ge 0,\, f_1(1) \le 0$, so $\exists x_0,\, f_1(x_0)=0 \, \Rightarrow f(x_0)=x_0$. Now let get the least $n$ for which $f^n(x)=x,\, \forall x\in [0,1]$. Thus $\exists x_1,\, \{f^j(x_1)\}_{j=0}^{n-1}$ are $n$ different points. Let assume that $n \ge 3$. In this case $\exists i_1,\,i_2$ such that $f^{i_1}(x_1)$ is between $x_0$ and $f^{i_2}(x_0)$. Let denote $y_1=f^{i_1}(x_1), \,y_2=f^{i_2}(x_1) $. Because $f^j(x)$ is periodic by $ j$, there exists $ m \in \mathbb{N}, \, f^m(y_1)=y_2$. Now we can apply the Lemma with $f^m(x)$ instead of $\phi$ and get that there is $x$ for which $ \{f^{jm}(x) \}_{j=0}^{\infty}$ consists more then $n$ points, which is a contradiction. Now there remain 2 possibilities: $n=1, \, f(x)=x$ and $n=2$. Let $f(f(x))=x, \, \forall x\in [0,1]$. Let $b=f(0)$, and $x_0, \, f(x_0)=x_0$. If $b=0$ and there exists $x_1, \, f(x_1) \neq x_1$ we can get contradiction using the Lemma with $0$ or $x_0$ as a fixed point depending of the sign of $f(x_1)-x_1$. Thus $b >0$. The case $ b \le x_0$ is impossible with similar arguments. So $f: [0,x_0]\to[x_0,b]$ is a bijection. If $b <1$ then $\exists \epsilon >0,\, f(b+\epsilon) \in [0,b)$ which is imposible. Finally we get that all required functions are: 1. The identity $f(x)=x$ 2.For any given continuous bijection $\phi :[0,a] \to [a,1], \, 0<a<1,\, \phi(0)=1,\, \phi(a)=a$, we construct $f(x)=\phi(x), \, x \in [0,a]; \, f(x)=\phi^{-1}(x),\, x \in (a, 1]$.

Not very different from the above solution. $f(a) = f(b) \Rightarrow f^n(a)=f^n(b) \Rightarrow a=b \Rightarrow $ $f$ is one-to-one and onto. So $f$ is decreasing or increasing. (Otherwise, at least for two points $f$ will have same value) a. Let $f$ be decreasing. $f(0)=1$ and $f(1)=0$. The function should meet $y=x$ at a point $0<k<1$. So $f(k)=k$. Let $a<k$, $f(a)=b$ and $f(b)=c$. Suppose $f^2(a)=c>a$. Since $b>k>c>a$, we have $f(a)>f^2(a)>a$. Taking $f$ of each side of an inequality means that inequality will change its order direction. So $f^2(a)<f^3(a)<f(a)$. Also we know $a<f^2(a)$. So $a < f^2(a)<f^3(a)<f(a)$. After taking $f$, we have $a < f(a) < f^3(a)<f^4(a)<f^2(a)$. $f^n(a)$ will always be between $f(a)$ and $f^2(a)$. So $f^n(a) \neq a$ for $c>a$. Suppose $f^2(a)=c<a$. Since $b>k>a>c$, we have $f(a)>a>f^2(a)$. So $f^2(a)<f(a)<f^3(a)$. Also we know $f^2(a)<a<f(a)$. $f^2(a)<a<f(a)<f^3(a)$. $f^3(a)<f(a)<f^2(a)<f^4(a) \Rightarrow f^3(a)<f(a)< a < f^2(a)<f^4(a)$. So $f^n \not\in \left[f(a), f^2(a)\right]$. Also $a\in \left(f(a), f^2(a)\right)$. So $f^n(a) \neq a$ for $c<a$. So $f^2(a) = c = a$. This means $f^2(x)=x$ for all $x$. For $0<k<1$, let $g: [0,k] \rightarrow [k,1]$, $g(0)=1$, $g(k)=k$, and $g$ is decreasing. \[ f(x) = \left\{ \begin{array}{ll} g(x) & \quad 0 \leq x \leq k \\ g^{-1}(x) & \quad k < x \leq 1 \end{array} \right. \] b. Let $f$ be increasing. $f(0)=0$ and $f(1)=1$. $f(a)>a \Rightarrow f(f(a))>f(a)>a \Rightarrow f^n(a)>a$. $f(a)<a \Rightarrow f(f(a))<f(a)<a \Rightarrow f^n(a)<a$. So $f(a)=a$. So the only increasing $f$ is $f(x)=x$.

Here is a similar, but somehow faster way to argue in the case when $f$ is decreasing. Notice that if $f$ is decreasing, $f^{2k}$ is increasing; and, $f^{2k+1}$ is decreasing. Since $f^n(x)=x$, and the RHS is increasing, we deduce that $2\mid n$. Now, supposing $\exists x_0\in [0,1]$ such that $f(f(x_0)>x_0$, we apply $f\circ f$ to this, and arrive at $f^4(x_0)>f^2(x_0)>x_0$; and continuing this (keep in mind that $n$ is even), we will eventually reach $f^n(x_0)>x_0$, a contradiction. Similar holds for $f(f(x_0))<x_0$, thus, $f(f(x))=x$ must hold for every $x$, in the case when $f$ is decreasing.