Basis of $\{f\in Map(\mathbb{F}_5, \mathbb{F}_5) | \sum_{i=0}^{4}f(i)=0\}$ I tried to find basis of $V=\{f\in Map(\mathbb{F}_5, \mathbb{F}_5) | \sum_{i=0}^{4}f(i)=0\}$ which is a subspace of $\mathbb{F}_5$ - vectorspace $Map(\mathbb{F}_5, \mathbb{F}_5)$.
What I tried:
I found this article and tried to create basis from basis of  $Map(\mathbb{F}_5, \mathbb{F}_5)$.
Also, I thought that $f\in Map(\mathbb{F}_5, \mathbb{F}_5)$ can be regarded as a vector $v_f=(f(0),f(1),f(2),f(3),f(4))\in \mathbb{F}_5^5$. But I don't know what to do with this idea.
I am not a native speaker so if there is weird English please tell me. 
Thank you in advance. 
 A: Since as you said
$$\mathrm{Map}(\mathbb{F}_5,\mathbb{F}_5)\ni f \mapsto \left(f(0), f(1), f(2), f(3), f(4)\right)\in \mathbb{F}_5^5$$
is an isomorphism of $\mathbb{F}_5$-linear spaces, it suffices to find a basis of the linear space of solutions of the equation
$$x_0+x_1+x_2+x_3+x_4 = 0$$
in $\mathbb{F}_5$. There is a standard procedure in linear algebra of finding a basis of a system of homogeneous equations. Let us apply it here. First write equation as
$$x_4 = -x_0 -x_1 - x_2 - x_3$$
Hence if $(x_0,x_1,x_2,x_3,x_4)$ is a solution, then
$$(x_0, x_1, x_2, x_3, -x_0 -x_1 - x_2 - x_3) =$$
$$= x_0\cdot (1,0, 0, 0, -1) + x_1\cdot (0, 1, 0, 0, -1) + x_2 \cdot (0,0, 1, 0, -1) + x_3\cdot (0,0,0,1,-1)$$
This implies that you can pick 
$$\{(1,0, 0, 0, -1),(0, 1, 0, 0, -1),(0,0, 1, 0, -1),(0,0,0,1,-1)\}$$ 
as a basis of this vector subspace. Now you can translate this vectors into functions in $\mathrm{Map}(\mathbb{F}_5, \mathbb{F}_5)$ by obvious inverse of the isomorphism you have given above.
