consider $A=\{a_1<a_2<....<a_k\}$. The main idea is to prove that $a_1$ isn’t negative. If $a_1$ is negative then $-a_1=b$ so b is positive, now since the $2$ sets are equal then there exists a number $x$ such that $f^b(x)-x=b+a_1=0$ so $f^b(x)=x$ so for that x we have $f^{2b}(x)=f^b(x)=x$. And we also know $f^{2b}(x)-x=2b+a_i$ so $a_i=-2b<-b=a_1$ this contradicts the minimality of $a_1$. So there are no negative numbers in $A$. The next part is actually very easy. Considering the largest or the lowest number between the positive integers, you can see that there are no positive numbers in $A$. {which I will add later} so the answer is the set $\{0\}$ and $f(x)=x+1$ works.

First of all a Solution : $\textbf{Claim 1 :-}$ $a_1,\dots,a_k \geq 0$ Proof: FtSoC suppose $a_1$ is the minimum $a_1 <0$ and let $a=-a_1$. We have for some $x$, $f^a(x)=x+a+a_1=x \implies f^a(x)=x$ So for some $a_j$ we have $x+2a+a_j=f^{2a}(x)=f^a(f^a(x))=f^a(x)=x \implies -2a=a_j \implies 2a_1=a_j$ but we had $a_1<0$ is the minimum. Absurd. Let $M>0$ be the maximum number in $\{a_1,\dots,a_k\}$ So there exists some $x$ s.t $f(x)=x+M+1$ Then $f^2(x)=x+2+a_i$ for some $i$ and also $f^2(x)=f(x+M+1)=x+M+2+a_j$ for some $j$ but then $a_j+M=a_i$ and we had $M$ is max and $a_1,\dots ,a_k \geq 0$ so $a_j=0 , a_i=M$ So $f^2(x)=x+M+2$ Similarly if for $i=1,\dots,l$ $f^l(x)=x+M+l$ then $f^{l+1}(x)=f(f^l(x))=f(x+M+l)=x+M+l+1+a_i$ Also $f^{l+1}(x)=x+l+1+a_j$ So we get $M+l+1+a_i=l+1+a_j \implies M+a_i=a_j \implies a_i=0 ,a_j=M$ So $\forall j$ $f^j(x)=x+j+M$ Let $f(x+M)=x+M+1+a_i \implies f^{a_i+1}(x)=f(x+M)$ If $a_i>0$, $f^{a_i}(x)=x+M \implies x+a_i+M=x+M \implies a_i=0$ Absurd. If $a_i=0$, $f(x+M)=x+M+1=f(x) \implies x=x+M \implies M=0$ Absurd. It means we should have $M=0$ so the only possible set is $\{0 \}$ .We need $f^i(x)-x=i$, notice $f(x)=x+1$ works fine. $\boxed{\mathcal{QED}}$ ${\color{red} \rule{25cm}{1pt}}$ A possible approach: $\textbf{Claim :-}$ $0 \in \{a_1,\dots,a_k\}$ Proof: $f^i(x)=x+i+a_t$ for some $t$. if we vary $i$ by PhP the are $i > j$ big enough such that for some $t$, $f^i(x)=x+i+a_t$ and $f^j(x)=x+j+a_t$ So $x+i+a_t=f^i(x)=f^{i-j}(f^j(x))=f^{i-j}(x+j+a_t)=x+j+a_t+i-j+a_u=x+i+a_t+a_u$ for some $u\in \{1,\dots,k \}$ Then $x+i+a_t=x+i+a_t+a_u \implies 0=a_u$ $\heartsuit$ Can someone finish this approach? [essentially in a way that the Claim becomes neccesarry]

if we remove the assertion "injectivity" then the folowing function works for every $k$tuple with $a_1=0$ and $a_i >0$ for others. $f(\frac{i}{k-1})=\frac{i}{k-1}+a_{i+1}+1$ for $0\le i \le k-1$ and $f(x)=x+1$ for every other $x$ . also your unfinished aproach on above is my solution on the exam. and it works

let $A = {a_1,..., a_n}$ be the set of numbers and $a_1<...<a_n$ lemma : $0 \le a_1$ Proof : if $0 > a_1$ then $a_1=-b$ and $b$ is positive , so exist $x$ such that: \[f^b(x)-x = b+a_1 = 0 \Rightarrow f^{2b}(x) - x = 0 = 2b+a_j \Rightarrow a_1\]so it isn't the smallest. now $a_i$ s are non negative , we again let $f^2(x)-x = 2+a_1$ but $f^2(x) - f(x) = 1+a_i \Rightarrow f(x) - x = 2 + a_i + a_j \Rightarrow  a_1 = a_i + a_j $ but $a_1$ is the smallest number and are non negative so $a_1 = 0$ now let $f(c) - c = a_n + 1$ and $b_i = f^i(c) - f^{i-1}(c)$ then $b_1 = a_n + 1$ and for $i > 1: b_i = 1$ because $a_n$ is the biggest number and we can check $f^s(c)-c$ now $f(c+a_n)$ equal $f^j(c)$ and for $i > 1: f^i(c) = i+c+a_n \Rightarrow f(c+a_n) = f^j(c) \Rightarrow c+a_n = j+c+a_n - 1 \Rightarrow j=1 \Rightarrow  c = c+a_n \Rightarrow a_n = 0 \Rightarrow A={0} , f(x) = x+1$

Proposed by Hesam Rajabzadeh.

Wrong....

I don't know but I don't think it's a "nice" problem.