I am going to assume that you meant to state that this was a faithful representation, meaning that the map $G \to GL_n(\mathbb{C})$ has no kernel. Otherwise, as Akhil says, the problem is dramatically underconstrained.
An infinite class of examples is the semidriect product of $S_5$ acting on $(\mathbb{Z}/N)^4$, for $N$ relatively prime to $30$. In other words, look at all permutation matrices whose nonzero entries are $N$-th roots of unity, with product $1$.
Here is another infinite family of examples: Let $G$ be any subgroup of $GL_5(\mathbb{C})$. Let $A$ be an abelian cyclic group, embedded in $GL_5(\mathbb{C})$ as a subgroup of the scalar matrices. Then $A$ is central in $GL_5(\mathbb{C})$, so $GA$ is a finite subgroup of $GL_5(\mathbb{C})$. If $\mathbb{C}^5$ is an irreducible representation of $G$, it is also an irreducible representation of $GA$.
Roughly speaking, the rest of my answer says that every example looks like one of these.
There is a theorem of Jordan which states that there is a bound $J(n)$ such that any subgroup $G$ of $GL_n(\mathbb{C})$ has a normal abelian subgroup $A$ with $|G/A| \leq J(n)$. See Terry Tao's blog for a nice exposition.
Using modern improvements on Jordan's result, we can get some more precise bounds.
Specifically, every $G$ as in your question is either:
(1) An extension $0 \to A \to G \to H \to 0$, where $A$ is abelian and $H$ is a subgroup of $S_5$ or
(2) A central extension $0 \to A \to G \to H \to 0$ where $|H| \leq 25920$ and $A$ is cyclic.
I think if I knew more about group cohomology I could, in principle, turn this into an infinite but comprehensible, list of all groups that may occur.
A representation $V$ of $G$ is called primitive it is irreducible and not induced from a proper subgroup $H$ of $G$.
Case 1: $\mathbb{C}^5$ is an irreducible, but imprimitive representation, of $G$. Let $\mathbb{C}^5$ be induced from a representation $L$ of $B \subset G$. Then $(\dim L) |G/B| = 5$, so $L$ is one dimensional and $H$ has index $5$ in $G$. Let $A = \bigcap_{g \in G/B} g B g^{-1}$. So $A$ is normal and $G/A$ acts faithfully on the $5$ element set $G/B$, showing that $G/A$ is a subgroup of $S_5$. We must show that $A$ is abelian. Suppose not, and let $C$ be the commutator of $A$. Then $C$ is killed by the map to $GL(L)$, as $L$ is one dimensional. But $C$ is normal in $G$, so it is also killed by the map to $GL(\mathbb{C}^5)$. This contradicts that we are supposed to be dealing with a faithful representation.
Case 2: $\mathbb{C}^5$ is primitive.
According to a recent paper of Collins, if $G$ has a a primitive representation of dimension $5$, then $|G/Z(G)| \leq 25920$. (Here $Z(G)$ is the center of $G$.) We take $A=Z(G)$. Since $\mathbb{C}^5$ is irreducible, we know that $Z(G)$ acts by scalars. So $Z(G)$ is a subgroup of $\mathbb{C}^*$, and must be cyclic.
I couldn't find a list of all groups that could occur as $G/Z(G)$, but perhaps you can extract it from Collins' work.