Plugging $y=1$ yields $f^{x+1}(1)+f(f(x))=2f(x+1)$, call this assertion $Q(x)$. We prove by induction that $f(n)=f^n(1)$ for all $n$. $Q(1)$ immediately implies $f(f(1))=f(2)$. Now if $f(n)=f^n(1)$ for some $n \geq 2$, then $Q(n)$ leads to $$2f(n+1)=f(f(n))+f^{n+1}(n)=f(f^n(1))+f^{n+1}(1)=2f^{n+1}(1),$$as desired. Conversely, any function with $f^n(1)=f(n)$ is a solution, since both sides of the required identity are equal to $2f^{x+y}(1)$. Now $Q(n)$ also implies that $f(f(n))=f(n+1)$ for all $n$. On the other hand, for any such function it follows by induction that $f^n(1)=f(n)$ for all $n$, thus such a function is a solution. To sum up, the problem is equivalent to $\boxed{f(f(n))=f(n+1) \ \forall n \in \mathbb{N}}$. This functional equation seems to have a lot of solutions and I am not sure if there is an easy general form for them....

Aryan-23 wrote: Find all functions $f : \mathbb N \mapsto \mathbb N$ such that the following identity $$f^{x+1}(y)+f^{y+1}(x)=2f(x+y)$$holds for all $x,y \in \mathbb N$ Answer Let $x_0, x_1, \dots, x_{t-1}\in \mathbb{N}_{\ge N}$ be distinct integers. Then the general solution set is \[ f(x)= \begin{cases} x+1&\text{if } 0<x<N\\ x_i &\text{if }x\geq N \land x-N \equiv i \pmod t, \end{cases} \] Solution Let $P(x, y)$ denote the functional equation. Firstly I contend: Claim: The problem statement is equivalent to $$f^x(1)=f(x)$$ for all $x\in \mathbb{N}$ Proof. From $P(1, 1)$ we find $f^2(1)=f(2)$, so the statment is true. Now proceed by induction on $x$. $$P(x,1): 2f(x+1)=f^{x+1}(1)+f^2(x)=f^{x+1}(1)+f(f^x(1))=2f^{x+1}(1)\implies f^{x+1}(1)=f(x+1)$$ so the result holds. Conversly, notice that by using the above $$2f(x+y)=f^{x+y}(1)+f^{x+y}(1)=f^x(f^y(1))+f^y(f^x(1))=f^x(f(y))+f^y(f(x))=f^{x+1}(y)+f^{y+1}(x)$$ so the reverse implication is also true. \(\blacksquare\) We now want to solve $f^x(1)=f(x)$ over $\mathbb{N}$. To do so it is a good idea to focus on the orbit of $1$ (taken as a multiset) $$\mathcal{O}(1)=\{f^i(1)\}_{i\geq 1}=\{f(1), f(2), f(3) \dots\}$$ Consider the following two scenarios: Case 1: No two elements of $\mathcal{O}(1)$ are equal. This means $f$ is injective, so $$f^{x+1}(1)=f(x+1)\implies x+1=f^{x}(1)=f(x)$$ Which leads to the solution $f\equiv n+1$. Case 2: There is a minimal $N$ such that $f(N)=f(N+t)$ for some minimal $t$. By the nature of $\mathcal{O}(1)$, it must be periodic from $N$ on with period $t$. So if we let $x_i=f(N+i)$ where $0\leq i\leq t-1$, then for $x\geq N$ $$f(x)=x_i, x-N\equiv i \pmod t$$ From the minimality of $N$, all elements of $\{f(1), f(2), \dots f(N-1)\}$ are distinct so we may apply Case 1 to get $f(x)=x+1$ when $1\leq x\leq N-1$. Given $\{x_0, x_t\dots x_{t-1}\}\cap \{f(1), f(2), \dots f(N-1)\}=\emptyset$ we get $x_i\geq N$ for all $0\leq i< t$. Hence we arrive at the general solution exposed at the beginning, and it is straightforward to check that it works.