For fixed $n \in \mathbb{R}$, what are all the continuous functions that satisfy $nf(x) = f(nx)$? For fixed $n \in \mathbb{R}$, what are all the continuous functions that satisfy $nf(x) = f(nx)$? 
I thought it would just be functions of the form $f(x) = kx$ but, for example, in the $n=2$ case we have that $f(x) =  x \cos (\frac{2\pi  \ln x }{ \ln 2})$ also works.
 A: Let's start with $n$ positive but not $1$.  We can separate positive and negative $x$, so take any periodic continuous functions $g_1(x)$ and $g_2(x)$ with $g_1(x\pm 1)=g_1(x)$ and $g_2(x\pm 1)=g_2(x)$ - a possibility is they are both constant, while another is that they each have period $1$, and they can be the fame function 


*

*for $x \gt 0$ let $f(x)=x g_1({\log_n (x)})$, 

*for $x \lt 0$ let $f(x)=x g_2({\log_n (-x)})$, 

*and $f(0)=0$  


while for $n=1$ you simply have 


*

*$f(x)$ is any continuous function


and with $n=0$, almost as simply,


*

*$f(x)$ is any continuous function with $f(0)=0$


It gets slightly more complicated for negative $n \not = -1$.  Take a continuous function $h(x)$ on $[0,1]$ with $h(1)=-h(0)$ and extend it to $\mathbb R$ with $h(x\pm 1)=-h(x)$, so if $h$ is ever non-zero then it has period $2$ and $|h|$ has period $1$ 


*

*for $x \gt 0$ let $f(x)=x h({\log_{-n} (x)})$, 

*for $x \lt 0$ let $f(x)=x h(1+{\log_{-n} (-x)})$, 

*and $f(0)=0$  


while for $n=-1$ you simply have 


*

*$f(x)$ is any odd continuous function so $f(-x)=-f(x)$ and $f(0)=0$


In particular, $f(x)=0$ is a possible solution for every value of $n$
