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As explained in my comment, $f(0)=0$ because otherwise Sarkovski's theorem would give us other periodic orbits which, of course, won't visit $x=0$. We also know that $f(x)<x$ for $x>0$.

Decompose the open set $\{x: f(x)>0\}=\bigcup I_n$ into its components. Clearly, since each set $[0,a]$ is invariant, the zeros of $f$ must accumulate at $0$. On the other hand, I claim that the $I_n$ do not accumulate at $x=0$.

Indeed, if they did, we could start out with any $I_0$ and then $f(\overline{I_0})=[0,b_0]$ for some $b_0>0$. By assumption, $I_1\subseteq [0,b_0]$ for some $I_1$. Let $K_1=\{x\in \overline{I_0}: f(x)\in \overline{I_1}\}$. Since $0\notin\overline{I_n}$ for all $n$, the orbit of any $x\in K_1$ will not yet have reached zero after one iteration.

Continue in this style: $f(\overline{I_1})=[0,b_1]\supseteq I_2$ for some $I_2$. Let $K_2 =\{x\in K_1: f^2(x)\in \overline{I_2}\}$. The compact sets $K_n$ are nested, so $\bigcap K_n\not=\emptyset$, but if $x\in K_n$, then $f^k(x)\not= 0$ for $k\le n$, so this point never reaches zero.

It follows that $f=0$ on $[0,d]$ for some $d>0$, but then everything is clear because now $f(x)\le x-\delta$ for some fixed $\delta>0$ for $x\ge d$, and each iteration brings us closer to the set $[0,d]$ by at least $\delta$.

Yes, this implies that $f$ is nilpotent.

As explained in my comment, $f(0)=0$ because otherwise Sarkovski's theorem would give us other periodic orbits which, of course, won't visit $x=0$. We also know that $f(x)<x$ for $x>0$.

Decompose the open set $\{x: f(x)>0\}=\bigcup I_n$ into its connected components. Clearly, since each set $[0,a]$ is invariant, the zeros of $f$ must accumulate at $0$. On the other hand, I claim that the $I_n$ do not accumulate at $x=0$.

Indeed, if they did, we could start out with any $I_0$ and then $f(\overline{I_0})=[0,b_0]$ for some $b_0>0$. By assumption, $I_1\subseteq [0,b_0]$ for some $I_1$. Let $K_1=\{x\in \overline{I_0}: f(x)\in \overline{I_1}\}$. Since $0\notin\overline{I_n}$ for all $n$, the orbit of any $x\in K_1$ will not yet have reached zero after one iteration.

Continue in this style: $f(\overline{I_1})=[0,b_1]\supseteq I_2$ for some $I_2$. Let $K_2 =\{x\in K_1: f^2(x)\in \overline{I_2}\}$. The compact sets $K_n$ are nested, so $\bigcap K_n\not=\emptyset$, but if $x\in K_n$, then $f^k(x)\not= 0$ for $k\le n$, so this point never reaches zero.

It follows that $f=0$ on $[0,d]$ for some $d>0$, but then everything is clear because now $f(x)\le x-\delta$ for some fixed $\delta>0$ for $x\ge d$, and each iteration brings us closer to the set $[0,d]$ by at least $\delta$.

As explained in my comment, $f(0)=0$ because otherwise Sarkovski's theorem would give us other periodic orbits which, of course, won't visit $x=0$. We also know that $f(x)<x$ for $x>0$.

Decompose the open set $\{x: f(x)>0\}=\bigcup B_n$ into its components, with $B_n=B_{r_n}(a_n)=(a_n-r_n,a_n+r_n)$. I claim that the $B_n$ can't accumulate at $x=0$. Indeed, if they did, we could start out with any $B_0$ and then $f(\overline{B_0})=[0,b_0]$ for some $b_0>0$. By assumption, $B_1\subseteq [0,b_0]$ for some $B_1$. Let $K_1=\{x\in \overline{B_0}: f(x)\in \overline{B_{r_1/2}(a_1)}\}$.

Continue in this style: $f(\overline{B_1})=[0,b_1]\supseteq B_2$ for some $B_2$. Let $K_2 =\{x\in K_1: f^2(x)\in \overline{B_{r_2/2}(a_2)}\}$. The compact sets $K_n$ are nested, so $\bigcap K_n\not=\emptyset$, but if $x\in K_n$, then $f^k(x)\not= 0$ for $k\le n$.

It follows that $f=0$ on $[0,d]$ for some $d>0$, but then everything is clear because now $f(x)\le x-\delta$ for some fixed $\delta>0$ for $x\ge d$, and each iteration brings us closer to the set $[0,d]$ by at least $\delta$.

(It might actually be clearer to describe this informally: The sequence $f^n(x)$ is decreasing, and if we don't have $f\equiv 0$ near zero, then there is first an interval $I_1$ with $f\not=0$ on $I_1$, and then a subinterval $I_2$ with $f^2\not= 0$ on $I_2$ etc., because we never run out of non-zero values in the image $[0,d]$, no matter how many times we iterate.)

As explained in my comment, $f(0)=0$ because otherwise Sarkovski's theorem would give us other periodic orbits which, of course, won't visit $x=0$. We also know that $f(x)<x$ for $x>0$.

Decompose the open set $\{x: f(x)>0\}=\bigcup I_n$ into its components. Clearly, since each set $[0,a]$ is invariant, the zeros of $f$ must accumulate at $0$. On the other hand, I claim that the $I_n$ do not accumulate at $x=0$. 

Indeed, if they did, we could start out with any $I_0$ and then $f(\overline{I_0})=[0,b_0]$ for some $b_0>0$. By assumption, $I_1\subseteq [0,b_0]$ for some $I_1$. Let $K_1=\{x\in \overline{I_0}: f(x)\in \overline{I_1}\}$. Since $0\notin\overline{I_n}$ for all $n$, the orbit of any $x\in K_1$ will not yet have reached zero after one iteration.

Continue in this style: $f(\overline{I_1})=[0,b_1]\supseteq I_2$ for some $I_2$. Let $K_2 =\{x\in K_1: f^2(x)\in \overline{I_2}\}$. The compact sets $K_n$ are nested, so $\bigcap K_n\not=\emptyset$, but if $x\in K_n$, then $f^k(x)\not= 0$ for $k\le n$, so this point never reaches zero.

It follows that $f=0$ on $[0,d]$ for some $d>0$, but then everything is clear because now $f(x)\le x-\delta$ for some fixed $\delta>0$ for $x\ge d$, and each iteration brings us closer to the set $[0,d]$ by at least $\delta$.

As explained in my comment, $f(0)=0$ because otherwise Sarkovski's theorem would give us other periodic orbits which, of course, won't visit $x=0$. We also know that $f(x)<x$ for $x>0$.

Decompose the open set $\{x: f(x)>0\}=\bigcup B_n$ into its components, with $B_n=B_{r_n}(a_n)=(a_n-r_n,a_n+r_n)$. I claim that the $B_n$ can't accumulate at $x=0$. Indeed, if they did, we could start out with any $B_0$ and then $f(\overline{B_0})=[0,b_0]$ for some $b_0>0$. By assumption, $B_1\subseteq [0,b_0]$ for some $B_1$. Let $K_1=\{x\in \overline{B_0}: f(x)\in \overline{B_{r_1/2}(a_1)}\}$.

Continue in this style: $f(\overline{B_1})=[0,b_1]\supseteq B_2$ for some $B_2$. Let $K_2 =\{x\in K_1: f^2(x)\in \overline{B_{r_2/2}(a_2)}\}$. The compact sets $K_n$ are nested, so $\bigcap K_n\not=\emptyset$, but if $x\in K_n$, then $f^k(x)\not= 0$ for $k\le n$.

It follows that $f=0$ on $[0,d]$ for some $d>0$, but then everything is clear because now $f(x)\le x-\delta$ for some fixed $\delta>0$ for $x\ge d$, and each iteration brings us closer to the set $[0,d]$ by at least $\delta$.

As explained in my comment, $f(0)=0$ because otherwise Sarkovski's theorem would give us other periodic orbits which, of course, won't visit $x=0$. We also know that $f(x)<x$ for $x>0$.

Decompose the open set $\{x: f(x)>0\}=\bigcup B_n$ into its components, with $B_n=B_{r_n}(a_n)=(a_n-r_n,a_n+r_n)$. I claim that the $B_n$ can't accumulate at $x=0$. Indeed, if they did, we could start out with any $B_0$ and then $f(\overline{B_0})=[0,b_0]$ for some $b_0>0$. By assumption, $B_1\subseteq [0,b_0]$ for some $B_1$. Let $K_1=\{x\in \overline{B_0}: f(x)\in \overline{B_{r_1/2}(a_1)}\}$.

Continue in this style: $f(\overline{B_1})=[0,b_1]\supseteq B_2$ for some $B_2$. Let $K_2 =\{x\in K_1: f^2(x)\in \overline{B_{r_2/2}(a_2)}\}$. The compact sets $K_n$ are nested, so $\bigcap K_n\not=\emptyset$, but if $x\in K_n$, then $f^k(x)\not= 0$ for $k\le n$.

It follows that $f=0$ on $[0,d]$ for some $d>0$, but then everything is clear because now $f(x)\le x-\delta$ for some fixed $\delta>0$ for $x\ge d$, and each iteration brings us closer to the set $[0,d]$ by at least $\delta$.

(It might actually be clearer to describe this informally: The sequence $f^n(x)$ is decreasing, and if we don't have $f\equiv 0$ near zero, then there is first an interval $I_1$ with $f\not=0$ on $I_1$, and then a subinterval $I_2$ with $f^2\not= 0$ on $I_2$ etc., because we never run out of non-zero values in the image $[0,d]$, no matter how many times we iterate.)