Maps with every point being periodic 
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*Does there exists a characterization of continuous maps $f:[0,1]\rightarrow [0,1]$ with every point $x\in [0,1]$ being periodic (i.e. if for every $x\in [0,1]$ there exists $n\in\mathbb{N}$ such that $f^{n}(x)=x$)?

*What about those continuous maps $f:[0,1]\rightarrow [0,1]$ with every point being periodic and for which the set of all periods $\{Per(x):\ x\in [0,1]\}$ is bounded?
 A: Nice question.
I'll answer your second question first. If all periods are bounded by $N$, then $f^{N!}$ is the identity. Thus $f$ is a homeomorphism, as $f^{N!-1}$ is an inverse. In particular $f$ must be monotonic.
EDIT: In fact, this holds under the hypotheses of question 1 as well. If $f^n(x)=x$ and $f^m(y)=y$ then $f^{mn}(x)=x$ and $f^{mn}(y)=y$. In particular $f(x)=f(y)$ implies $x=y$. Thus $f$ is injective, so it must be monotonic. Thanks to @JonathanY. for this observation.
Suppose that $f$ is monotonically increasing, and suppose there is some $x$ such that $f(x)\neq x$. If $f(x)>x$ then $f^n(x)>x$ for all $n$. Similarly if $f(x)<x$ then $f^n(x)<x$ for all $x$. Thus we must have $f(x)=x$ for all $x$.
If on the other hand $f$ is monotonically decreasing then $f^2$ is monotonically increasing, so we must have $f(f(x))=x$ for all $x$. There are many such functions, but they all have this form: take any $0<\omega<1$ and any decreasing homeomorphism $g:[0,\omega]\to[\omega,1]$ and define (with obvious notation) $f=g\cup g^{-1}$.
EDIT: Every such $f$ is "homeomorphically-conjugate" to $1-x$, in the sense that there exists a homeomorphism $h:[0,1]\to[0,1]$ such that $hfh^{-1}(x)=1-x$. One such $h$ is
$$
 h(x) =
 \begin{cases}
   \frac{x}{2\omega} & \text{if $x\leq\omega$,}\\
   1 - \frac{f(x)}{2\omega} & \text{if $x\geq\omega$.}
 \end{cases}
$$
Indeed, then $h(f(x))=1-h(x)$. Thank you to @GerryMyerson for this observation.
[Old Baire argument, now defunct:
Now for your first question. Let $X$ be the set of all $x\in[0,1]$ for which there does not exist an integer $n$ and an open neighbourhood $U$ of $x$ such that $f^n(y)=y$ for all $y\in U$. Clearly $X$ is closed, so it is a complete metric space. Let $E_n$ be the set of all $x\in X$ such that $f^n(x)=x$. Then each $E_n$ is closed in $X$, and $X=\bigcup_{n=1}^\infty E_n$. If $X\neq\emptyset$, the Baire category theorem implies that some $E_n$ has nonempty interior, which leads straight to a contradiction. Thus $X=\emptyset$, and so by compactness of $[0,1]$ there must exist $N$ such that $f^N(x)=x$ for all $x$.]
