VMO

For $a=0$, we can check that the function $f(x)=2016$ for all real numbers $x$ is satisfied.

Now we consider the case $a\ne 0$. By plugging $x=-f(y)$ in condition ii), we have $$f(y)=f(-f(y))+ay$$ for all real numbers $y$. Hence, $f$ is injective.

Next, by letting $y=0$ in ii), we obtain $$f(x+f(0))=f(x)$$ for all real numbers $x$. Thus, $f(0)=0$.

Setting $y=\frac{-f(x)}{a}$ in ii) and combining the injectivity of $f$, we obtain that $$-\frac{f(x)}{a}+f\left(-\frac{f(x)}{a}\right)=-x$$ for all real numbers $x$.

Replacing $y$ by $-\frac{f(y)}{a}$ in ii) and applying the above equation, it implies $$f(x-y)=f(x)-f(y)$$ for all real numbers $x,y$. Thus, $f$ is additive and we can easily compute $f(2016)=2016f(1)=2016^2$.

On the other hand, because $f$ is additive then the condition ii) can be rewritten as $f(y)+f(f(y))=ay$ for all real numbers $y$. Letting $y=1$, we have $a=2016\cdot 2017$. For $a=2016\cdot 2017$, we can directly check that $f(x)=2016x$ is satisfied.

Therefore, $a=0$ or $a=2016\cdot 2017$ are desired values. ​