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

Now we consider the case $a \neq 0$. 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. $\boxed{}$