If $m > 2$ then the positive integers which are less than $m$ and co-prime to $m$ can be partitioned into pairs $(k, m-k)$, $m-k \neq k$. The number of these pairs is $\frac{\varphi(m)}{2}$ and the sum of the numbers in every pair is $m$. Therefore $f(m) = \frac{m\varphi(m)}{2}$, which is true for $m = 2$ as well. Since $\varphi(n^k) = n^{k-1}\varphi(n)$, we obtain
$$f(n^k) = \frac{n^k \varphi(n^k)}{2} = \frac{n^{2k-1} \varphi(n)}{2}.$$
Hence the given condition becomes $n^{2k-1}\varphi(n) = 2n^l$.
If $l > 2k-1$, then we obtain $\varphi(n) = 2n^{l-2k+1} \ge 2n$, which is impossible. If $l = 2k-1$ we get $\varphi(n) = 2$, which gives $n = 3, 4$ or $6$. If $l < 2k-1$, then $n^{2k-1-l}\varphi(n) = 2$, which is satisfied only by $n = 2$ and $2k-1-l = 1$, i.e. $l = 2k-2$.
Finally, the solutions are $n = 2, 3, 4$ and $6$.