, 2024

Answer: $f(x) \equiv 1$ and $f(x) = x$.
By putting $a = 1$ to $f(\varphi(a)) = \varphi(f(a))$ we get $f(1) = \varphi(f(1))$ and hence $f(1) = 1$. By putting $a = 2$ we get $f(1) = \varphi(f(2))$ and since $f(1) = 1$ we get that either $f(2)$ is either 1 or 2.

Case 1: $f(2) = 1$. Since $1 = f(2) = f(\varphi(3)) = \varphi(f(3))$ we get that $f(3)$ is either 1 or 2. But since $(3-1) \mid f(3) - f(1)$ we get that $f(3) = 1$. By induction we show that $f(n) = 1$ for all positive integers $n$. Assume that $f(a) = 1$ for all $a = 1, 2, \dots, n-1$. Then since $f(\varphi(n)) = \varphi(f(n))$ and $\varphi(n) \le n-1$ we get $\varphi(f(n)) = 1$ and hence $f(n)$ is either 1 or 2. Finally again since $n - (n-2) \mid f(n) - f(n-2)$ we get that $f(n) = 1$. Done.

Case 2: $f(2) = 2$. By induction we show that $f(n) = n$ for all positive integers $a$. Assume that $f(a) = a$ for all $a = 1, 2, \dots, n-1$. By the first condition and by inductive hypothesis for each $1 \le a \le n-1$ we have $n-a \mid f(n)-a$ or equivalently $n-a \mid f(n)-n$. Therefore, for each integer $1 \le m \le n-1$ we have $m \mid f(n)-n$. Hence the common divisor $M$ of all integers $1 \le m \le n-1$ also divides $f(n)-n$; $f(n) = Ms+n$, where $s$ is integer. Then since $\varphi(n) < n$ by the second condition and by inductive hypothesis $\varphi(n) = \varphi(Ms+n)$. On the other hand, since each integer $k$ with $(n, k) = 1$ and $k < n$ divides $M$ we have $(k, Ms+n) = 1$. Therefore, $\varphi(n) \le \varphi(Ms+n)$. If $s \ne 0$ then $(Ms+n-1, Ms+n) = 1$ and hence we get that $\varphi(n) < \varphi(Ms+n)$. This contradiction shows that $s=0$ and $f(n) = n$. Done.