Let $R$ be a ring, a $R$-module is called simple if it has no proper, nontrivial submodules. Let $G$ be a finite group, and denote by $\mathbb C[G]$ the free vector space over $G$, with the product given by multiplying the basis vectors according to the multiplication in $G$ and extending lineary. Alternatively this set could be thought of as the set of functions $G \to \mathbb C$ with the convolution product $$ (f \ast h)(x) = \sum_{y,z \in G, yz = x} f(y)h(z) = \sum_{y\in G} f(y)h(y^{-1}x). $$ Then $\mathbb C[G]$ is a ring. If $M$ is a module over $\mathbb C[G]$ which is simple, then $M$ is also a vector space over $\mathbb C \subseteq \mathbb C[G]$. Does the simplicity of $M$ as a $\mathbb C[G]$-module implies there is just one unique representation $\pi : G \to GL(V)$ (a representation is a homomorphism $G \to GL(V)$)?
As $\pi(g)v := g\cdot v$ (where as $G \subseteq \mathbb C[G]$ the product $g \cdot v$ is meant to be in the module) is one representation, this would be then the only one.
Suppose that $V$ is any finite dimensional $\mathbb{C}G$ module (I don't need to assume irreducible). Given any basis $\{v_1,\ldots,v_n\}$ I can write $$g.v_j=\sum_{i=1}^na_{ij}(g)v_i \;\;\;(j=1,\ldots,n)$$ for some scalars $a_{ij}(g)$. The assignment $\pi:G\to\mathrm{GL}(V)$, $g\mapsto (a_{ij}(g))_{i,j=1}^n$, defines a representation of $G$ and every representation can be obtained in this way.
The degree of freedom one has to define such a representation is a choice of basis. In particular, if $\pi,\pi':G\to\mathrm{GL}(V)$ are representations of the same $G$-module $V$, then there exists an invertible linear transformation $\phi:V\to V$ such that $$\phi\pi(g)=\pi'(g)\phi\;\;\;\mbox{for all }g\in G.$$ In other words, $\pi'(g)=\phi\pi(g)\phi^{-1}$ are conjugate matrices for every $g\in G$.
Added: Based on you're comment, there seems to be a need for a review of some linear algebra. Let's say $\{w_1,\ldots,w_n\}$ is another basis and $$g.w_j=\sum_ib_{ij}(g)w_i$$ so that $\pi'(g)=(b_{ij}(g))_{ij}$ is the corresponding representation. Since $\{v_i\}$ and $\{w_i\}$ are two bases, we can write $$v_j=\sum_{k}c_{kj}w_k$$ for some invertible matrix $(c_{ij})_{i,j}$. Then, at the level of $\mathbb{C}G$-modules, we have \begin{align*} g.v_j&=g.\left(\sum_k c_{kj}w_k\right)\\ &=\sum_{i,k}b_{ik}(g)c_{kj}w_i \end{align*} and \begin{align*} g.v_j&=\sum_ka_{kj}(g)v_k\\ &=\sum_ka_{kj}(g)v_k=\sum_{i,k}c_{ik}a_{kj}(g)w_i \end{align*} Therefore, taking the difference, $$0=\sum_i\left(\sum_k(b_{ik}(g)c_{kj}-c_{ik}a_{kj}(g))\right)w_i$$ Now, using the fact that $\{w_i\}$ is a basis, we have $$\sum_{k}c_{ik}a_{kj}(g)=\sum_{k}b_{ik}(g)c_{kj}$$ for all $i,j$.
Let $\phi:V\to V$ be the invertible linear transformation $\phi(v_i)=\sum_kc_{ki}w_k$. Then, the computation above shows that on the level of representations $\phi\pi(g)=\pi'(g)\phi$ as stated above.