Suppose $I$ is the set of vertices of the regular icosahedron, here is a link of the icosahedron: http://www.werheit.mynetcologne.de/icosaeder.gif
Let $F(I)$ be the space of complex functions on $I$, that is functions $f\colon I\rightarrow \mathbb{C}$. We know that $A_5$ acts on the icosahedron so we have a 12 dimensional representation of $A_5$: I believe that the representation is defined by $\rho(g) f(x)=f(gx)$ for $g\in A_5$. Now I have to decompose this representation in a direct sum of irreducible representations. Of course the function which is constant, gives a 1-dimensional sub representation.
What I thought was that if you choose for example 4 different complex numbers and you assign these 4 complex numbers to the vertices in all different ways, then you will have that this subspace will be invariant under the actions of $A_5$, because that is how you constructed it. I don't know if this is the way I should tackle this problem. Thanks.
It's hard to help without knowing your background knowledge.Here is one approach. The transitive action of $A_5$ of degree $12$ has stabilizer of order $5$, which has four orbits, two of size $5$ and two of size $1$. So $A_5$ has four orbits on $I^2$ so, by a standard result on permutation characters (the fixed point formula applied to $A_5$ on $I^2$), we have $\langle \chi,\chi \rangle = 4$, where $\chi$ is the character of $f$.
Now the degrees of the irreducible characters of $A_5$ are $1,3,3,4, 5$, so the only possibility for the decomposition giving $\langle \chi,\chi \rangle = 4$ is as $12 = 1+3+3+5$.