Integers and integer functions

I do not know how to define all solutions for all values, but I know how to find all solutions on $\{0,1,\dots,n\}$.

Depending on the choice of the parameters, there are up to $4$ possible solutions in $\{0,1,\dots,n\}$.

Defining $f(0)$

If all $a$s $= 0$, we have $f(0)=\frac{n}{k}f^n(0)$, thus either $f(0)=0$ or $f(0)=\left(\frac{k}{n}\right)^{\frac{1}{n-1}}$ if this number happens to be integer. For your numerical example, this number is not an integer.

Defining $f(1)$ given $f(0)$

If $a_1=1$ and the rest is $0$, we have $f(1)=\frac{n-1}{k}f^n(0)+\frac{1}{k}f^n(1)=\frac{n-1}{n}f(0)+\frac{1}{k}f^n(1)$. If $f(0)=0$, we have either $f(1)=0$ or $f(1)=k^{1/(n-1)}$; for your numerical example, the latter value is $2014$. If $f(0)\neq 0$, we can have between $0$ and $2$ possible values for $f(1)$ - we can use them as long as they are integer.

Defining $f(2), \dots, f(n)$ given $f(1)$ and$f(0)$

Once we have decided on the values for $f(0)$ and $f(1)$, we can define $f(m)$, $1<m\leq n$, by choosing $a_1=\dots=a_m=1$, and the rest is $0$.

We then have:

$f(m)=\frac{1}{k}\left(m f^n(1) + (n-m) f^n(1)\right) = \frac{1}{k}\left(m f^n(1) + (n-m) f^n(0)\right)=m\left(f(1)-\frac{n-1}{n}f(0)\right) + \frac{n-m}{n}f(0)=mf(1)+(1-m)f(0)$

This holds for any $f(1)$ and $f(0)$.

For your parameters, we have $f(m)=2014m$ and $f(m)=0$