Thanks very much, you're right, it is not true for $\mu=1/4$, meaning that in general the real zeros of the function $f(x)$ are not just $\mu^{2n}$, $n$ an integer. I thought that this perhaps was the case, implying what I wanted to prove. I'll explain more:
Define $$\chi_q(t):=t^2\int_0^\infty \mu^{-4\langle\log_{\mu^4}(s)+q\rangle }s e^{-(\mu^{-1}-\mu) ts}ds ,\quad \Re(t)>0$$ where $\mu$ and $q$ are constants with $0<\mu<1$, $0\leq q < 1$, and $\langle x\rangle$ denotes the fractional part of $x$.
The function I am looking at originally is $$g_q(t)=\chi_q(t)-\chi_q(\mu^2/t),\quad \Re(t)>0.$$ This functions satisfies $g_q(\mu^4t)=g_q(t)$. What I thought would be the case is that $g_q(t)$ has $\mu$ as its only zero in the region $\Re(t)>\mu^2R/2$, $|t-1/R|<1/R$. Here $R=\mu^{-1}+\mu$. This region is its own reflection about the circle of radius $\mu$, and the intersection of this region with the real axis is the interval $(\mu^2R/2, 2/R)$.
The function $s\mu^{-4\langle\log_{\mu^4}(s)+q\rangle }$ is a step function, and by integrating we can write $g_q(t)$ as the doubly infinite series (up to a multiplicative constant):$$t\sum_{k=-\infty}^\infty\mu^{4(k-q)}e^{-(\mu^{-1}-\mu)\mu^{4(k-q)}t}-\frac{\mu^2}{t}\sum_{k=-\infty}^\infty\mu^{4(k-q)}e^{-(\mu^{-1}-\mu)\mu^{4(k-q)}(\mu^2/t)}$$
For $q=0$, this is essentially the function in the original question since $f(x)=\mu^{-1} g_0(\mu x)$.
However, computational experiments with Mathematica indicate that for small values of $\mu$, g_0(t) has at least three real zeros in the interval $(\mu^2R/2, 2/R)$, and this seems to be the case as well for $\mu=1/5$ and $q=1/1000$, which disproves my conjectureat least for $q=0$. Interesting though, when I give $q$ values $>0$, the same experiments point to $\mu$ as the only zero in that interval. Below is the Mathematica code. Here m is $\mu$, and I truncate the series at $k=M$.
M = 300;
m = 1/5;
q = 0;
N[m]
R = m + 1/m;
g[x_] := x*(Sum[ m^(4*(k - q))Exp[-(1/m - m) m^(4*(k - q))x], {k, 0, M}] + Sum[m^(4(-k - q))Exp[-(1/m - m) m^(4*(-k - q))(m^2/x)], {k, 1, M}]) - (m^2/ x)(Sum[m^(4*(k - q))*Exp[-(1/m - m)m^(4(k - q))(m^2/x)], {k, 0, M}] + Sum[m^(4(-k - q))Exp[-(1/m - m) m^(4*(-k - q))*(m^2/x)], {k, 1, M}]);
Plot[g[x], {x, m^2*R/2, 2/R}, PlotRange -> All]
