$f(x)=1$ is not the only function satisfying the given relation.
Define an auxiliary function $g(x)=f(5^x)$, then the given relation is
and the derived equations are
From the last two equations above we can get
(corresponding to $f(x)=f(1/x)$). From equations $(1,2,3)$ we can derive the original relation and conditions, so we lose nothing by using only them.
It is clear that $g(x)=1$ for all integer $x$, but we can set $g(x)=-1$ for all non-integer $x$ and still satisfy all relations. In general, we can have for an arbitrary $n\in\Bbb N^+$:
and the relations would still be satisfied. Because $\frac1n\Bbb Z$ (the set of numbers whose product with $n$ is an integer) is closed under addition and subtraction, $g(x)=1$ is only fixed for those $x$ that are members of that set, leaving us free to map non-members to 1. Of course, $g(x)=1$ for all $x$ is also a solution.
Translated back, $f(x)$ is either constant 1 (the given solution) or
where $n\in\Bbb N^+$.