$f(nx)=f(x), \qquad n \in \mathbb Z^+$

Let $f$ satisfy the following equation, $$f(nx)=f(x), \qquad n\text{ a positive integer}.$$ Then I know that the most general solution is $$f(x)= C_{+}x^{2\pi i m/\log n}+C_{-}x^{-2\pi i m/\log n}.$$ However the solution will depend of two integers $m$ and $n$.

My question is what extra condition can I impose to my solution , (apart from the invariance under dilations) so the general solutions depend only on the number $n$ so $$f(x)= C_{+}x^{2\pi i n/\log n}+C_{-}x^{-2\pi i n/\log n}.$$

as an extra information the function f(X) is the solution to the eigenvalue equation $x \frac{df}{dx}=kf(x)$ here 'k' is a (complex) eigenvalue.

Are there hidden conditions on $f$ that you're not showing? Otherwise, you can take any function $g:\{a+bi\mid 0\le a<\log n, -\pi<b\le\pi\}\to\mathbb C$, extend it periodically to all of $\mathbb C$ and set $f(z) = g(\mathrm{Log}\;z)$. –  Henning Makholm Oct 22 '11 at 20:43
oh sorry i forgot... i wanted to solve the differential equation $x\frac{df}{dx}=kf(x)$ for an eigenvalue 'k' with the boundary conditions $f(nx=f(x)$ for any integer 'n'. HOwever my solution depend on 2 integers m and n i would like to throw the dependence on m so the eigenvalues are $k_{n} = \frac{2\pi n}{logn}$ –  Jose Garcia Oct 23 '11 at 9:27
"$f(nx)=f(x)$ for any integer $n$"? This seems to imply $f(qx)=f(x)$ for any (positive?) rational $q$ –  Henry Oct 24 '11 at 10:12