jmc

Note that $2^4=4^2,$ so from (iii), either $f(2)=4$ or $f(4)=2.$ But from (i), $$f(4)>f(3)>f(2)>f(1),$$so $f(4)\ge 4.$ Hence, $f(2)=4.$ By applying (ii) repeatedly, we find that $$f(2^n)=2^{2n}$$for all positive integers $n.$

From (i) and (iii), $$f(3{)}^2=f(9)>f(8)=64,$$so $f(3)\ge 9.$

Similarly, $$f(3{)}^8=f(3^8)<f(2^{13})=2^{26},$$so $f(3)\le 9.$ Therefore, $f(3)=9.$ It follows that $f(3^n)=3^{2n}$ for all positive integers $n.$

Now, $$f(5{)}^3=f(5^3)<f(2^7)=2^{14},$$so $f(5)\le 25.$

Also, $$f(5{)}^{11}=f(5^{11})>f(3^{16})=3^{32},$$so $f(5)\ge 25.$ Therefore, $f(5)=25.$

Hence, $$f(30)=f(2)f(3)f(5)=4\cdot 9\cdot 25=\boxed{900}.$$Note that the function $f(n)=n^2$ satisfies all the given properties. (It can be shown that the only solutions to $n^m=m^n$ where $m\ne n$ are $(2,4)$ and $(4,2).$)