We restrict attention to the set
$$S=\left\{1, f(1), f^{2}(1), \ldots\right\}$$
Observe that $S$ is unbounded because for every number $n$ in $S$ there exists a $k>0$ such that $f^{2 k}(n)=n+k$ is in $S$. Clearly $f$ maps $S$ into itself; moreover $f$ is injective on $S$. Indeed if $f^{i}(1)=f^{j}(1)$ with $i \neq j$ then the values $f^{m}(1)$ start repeating periodically from some point on, and $S$ would be finite. Define $g: S \rightarrow S$ by $g(n)=f^{2 k_{n}}(n)=n+k_{n}$. We prove that $g$ is injective too. Suppose that $g(a)=g(b)$ with $a \neq b$ and $k_{a} > k_{b}$. So, since $f$ is injective on $S$, we obtain
$$f^{2\left(k_{a}-k_{b}\right)}(a)=b=a+\left(k_{a}-k_{b}\right) .$$
However this contradicts the minimality of $k_{a}$ as $0 < k_{a} - k_{b} < k_{a}$. Therefore, $g$ is injective.

Let $T=\{n \in S : k_{n} \leq n\}$. Since $k_{n} > n$ for $n \in S$, $T$ is non-empty. For each $t \in T$ denote $C_{t}=\left\{t, g(t), g^{2}(t), \ldots\right\}$; call $C_{t}$ the chain starting at $t$. Observe that distinct chains are disjoint because $g$ is injective. Each $n \in S \backslash T$ has the form $n=g\left(n^{\prime}\right)$ with $n^{\prime} < n$, and $k_{n} > k_{n^{\prime}}$. Since $n^{\prime} \in S \backslash T$, we have $k_{n^{\prime}} > n^{\prime}$, and thus $k_{n} > n^{\prime} + 1 \geq n$. In conclusion $k_{1}, k_{2}, \ldots$ is unbounded.