The following is from the book Algebra by P.M. Cohn. Let $V$ be a $G$-module of dimension $d$ with basis $\{ v_1,...,v_d\}$. Then there is a corresponding representation $\rho : G\rightarrow\text{GL}_d$ by defining the action of $x\in G$ on $v\in V$ by
$$ v_ix=\sum_j\rho_{ij}(x)v_j $$ The author says that the representation defined in this way is "afforded" by the $G$-module. If we take another basis $u=(u_1,...,u_d)$, then we again get a representation $\sigma$ by defining $$ u_ix=\sum_j\rho_{ij}(x)u_j $$ Let $P$ be the matrix of transformation from $u$ to $v$ so that $$ v=Pu,\qquad u=P^{-1}v. $$ and therefore $$ \sigma(x) u=u x=\left(P^{-1} v\right) x=P^{-1}(v x)=P^{-1} \rho(x) v=P^{-1} \rho(x) P u . $$ which shows that $\sigma(x)=P^{-1}\rho(x)P$, and hence $\sigma$ and $\rho$ are equivalent representations.
The above is easy to understand. But the author then writes "Moreover, since $P$ may be any invertible matrix, we see that representations of $G$ that are equivalent are afforded by the same $G$-module for suitable bases" - this part I do not understand. How exactly do we see from the above that the $G$-module $V$ affords all representations that are equivalent to $\rho$?