Find all continuous functions defined over real numbers that satisfy
$\frac{f(x)}{f(y)} = \frac{f(kx)}{f(ky)}$,
for any $x$ and $y$. It is possible to show that the above condition holds for $f(x) = ax^b$ since
$\frac{ax^b}{ay^b} = \frac{ak^bx^b}{ak^by^b}$.
Do functions that satisfy this property have a specific name?
I will assume that $f$ never vanishes and $k>0$. Let $g(x)=\frac {f(kx)} {f(x)}$. The given equation becomes $g(x)=g(y)$ for all $x,y$ so $g$ is a constant $c$. Thus $f(kx)=cf(x)$ for all $x$. Put $x=0$ to see that $c=1$. Let $h(x)=f(e^{x})$. Then $h(x+p)=h(x)$ for all $x$ where $p=\log \, k$. You can retrace these steps and show that any function $h$ with $h(x+p)=h(x)$ (i.e. any periodic function $h$ with period $p$ which never vanishes) gives a solution to the given problem.
NOTE: There are discontinuous functions satisfying the given identity.
If $f$ is strictly monotonic (see comment by OP below) then $f(kx)<f(x)$ (for $x >0 0$) if $ k<1$ and $f(kx)>f(x)$ (for $x > 0$) if $ k>1$. Hence there is no solution unless $k=1$. Of course, the question is trivial when $k=1$.