A strictly increasing f with f(f(n)) = 3n

Step 1: $f(3n)=3f(n)$. $f(3n)=f(f(f(n)))=3f(n)$ (apply the equation to $f(n)$).

Step 2: $n<f(n)<3n$. If $f(n)\le n$, monotonicity gives $3n=f(f(n))\le f(n)\le n$, impossible. If $f(n)\ge3n$, then $3n=f(f(n))\ge f(3n)=3f(n)\ge9n$, impossible.

Step 3: small values. $1<f(1)<3$ so $f(1)=2$, and $f(2)=f(f(1))=3$. By Step 1, $f(3^k)=2\cdot3^k$ and $f(2\cdot3^k)=3^{k+1}$.

Step 4: first half of each block. The integers $3^k,\dots,2\cdot3^k$ are $3^k+1$ numbers, and $f$ maps them strictly increasingly into $\{2\cdot3^k,\dots,3^{k+1}\}$, which also has $3^k+1$ elements, with the end points matching. So the map is the shift:
$$f(3^k+j)=2\cdot3^k+j,\qquad0\le j\le3^k.$$

Step 5: second half. For $0\le j\le3^k$, $f(2\cdot3^k+j)=f\bigl(f(3^k+j)\bigr)=3(3^k+j)=3^{k+1}+3j$.

Formula. For $3^k\le n\le2\cdot3^k$: $f(n)=n+3^k$. For $2\cdot3^k\le n\le3^{k+1}$: $f(n)=3n-3^{k+1}$. (The two formulas agree at the common point $n=2\cdot3^k$: $f=3^{k+1}$ in both. Both pieces are increasing and they match at the junction, so $f$ is strictly increasing.)

Verification. For $n$ in the first range, $f(n)=n+3^k$ lies in the second range, $f(f(n))=3(n+3^k)-3^{k+1}=3n$ ✓. For $n$ in the second range, $f(n)=3n-3^{k+1}$ lies in $[3^{k+1},2\cdot3^{k+1}]$ and $f(f(n))=3n-3^{k+1}+3^{k+1}=3n$ ✓.

The value. $3^6=729$, $2\cdot3^6=1458$, $3^7=2187$. Since $1458\le2024\le2187$ we use the second formula with $k=6$:
$$f(2024)=3\cdot2024-2187=3885.$$

Lesson. For $f(f(n))=g(n)$ with $f$ monotone, trap $f$ between two bounds, find the values on a geometric sequence, and use the pigeonhole/order argument to fill in the blocks.