Application of Liouville’s Theorem Let $f$ : C → C be an entire function such that $2f(1) = f(0)$. What can you say about $f$?
Since $f(1) - f(0) =-f(1)$, My intuition says to define a new function $g(z)$ such that:
$g(z) = 
 \begin{cases} 
      f'(0) & z= 0 \\
      f(z)-f(0) &  z\neq 0  
   \end{cases}
$. 
But this is not helping me anyway to prove that $f$ is constant.
Edit: $f$ need not be constant as $f(z) = z(z-1)$ act as a counter example. 

Since $f$ is entire has everywhere (absolutly) convergent power series representation$f(z)=a_0+a_1z+a_2z^2+\cdots.$ According to your given condition $a_0=2(a_0+a_1+a_2+\cdots).$ Therefore $$f(z)=a_1(z-2)+a_2(z^2-2)+a_3(z^3-2)+\cdots.$$
For example:
$a_1=1-, a_2=1$ and $a_3=a_4=\cdots=0$ gives the function $z(z-1).$
$a_n=\left(\dfrac{\ln(1/2)}{n!}\right)^n\,\,\,\,\forall n=1,2,\cdots$ gives us the function $2^{-z}.$ 

In fact for any entire function $F,$ $$f(z)=F(z)-2F(1)+F(0)$$ has this property. 

