For which $a$, the DE $f'(x)=f(ax)$ has non- zero solution The question is 


  
    For which $a$, the differential equation $$f'(x)=f(ax)$$ has non- zero solution(s)?
  


Of course, for $a=1$, we have $f'(x)=f(x)$ which has the solution $f(x)=ce^{x}$.

For $a=-1$, the DE is $f'(x)=f(-x)$ which gives
$$f''(x)=-f'(-x)=-f(x)$$
and this gives $$f(x)=c_1\cos x+c_2\sin x$$
where from $f'(x)=f(-x)$, we see $f'(0)=f(0)$ and thus for $a=-1$, the solution is
$$f(x)=c_1(\cos x+\sin  x).$$
For $a\ne\pm1$, I have no idea how to proceed. 

Thanks for helps.
You may did as for $f'(x)=f(-x)$ for general case $f'(x)=f(ax)$.

Note that from $f'(x)=f(ax)$ we have
    $$f''(x)=af'(ax)=af(a^2x),\quad f'''(x)=a^3f'(a^2x)=a^3f(a^4x), \ldots$$
    and in general
    $$f^{(n)}(x)=a^{n(n-1)/2}f(a^nx)$$
    which gives $$f^{(n)}(0)=a^{n(n-1)/2}f(0).$$
    Therefore the $f(x)$ is given by
    $$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n=f(0)(1+\sum_{n=1}^\infty \frac{a^{n(n-1)/2}}{n!}x^n)$$
    and by ratio test, we see that $$|a|\leq 1.$$
Hint. Let $y=f(x),\,w=ax,$ then we get $y'_x=y'_ww'_x=y(ax),$ which gives $$ay'=y,$$ where $y$ is now a function of $w.$ You should now be able to proceed.