An algebra problem by Nico Stirling

As the domain and range only allows natural numbers, we plug in $n=1, 2, 3$ to get $f(f(1))=3$ , $f(f(2))=6$ and $f(f(3))=9$ . This gives $3$ possibilities.

Case $1$ : If $f(1)=1$ , then $f(f(1))=1\neq3$ . Therefore, no solutions here.

Case $2$ : If $f(1)=2$ , then $f(f(1))=f(2)=3$ . This gives $f(3)=6$ . So, we can obtain $f(6)=9$ and $f(9)=18$ . Given the strict inequality that the function is increasing, we have $f(4)=7, f(5)=8$ So, we can get $f(f(4))=f(7)=12$ and $f(f(7))=f(12)=21$ . Since $18\leq f(x)\leq21, x\in[9, 12]$ It follows that $f(10)=\boxed{19}$ .

Note that there is a third case whereby $f(1)=3$ . However, it contradicts $f(f(3)=9$ hence no solutions here.