The definition of equivalent representations is given by:
Definition: Let $G$ be a group, $\rho : G\rightarrow GL(V)$ and $\rho' : G\rightarrow GL(V')$ be two representations of G. We say that $\rho$ and $\rho'$ are $equivalent$ (or isomorphic) if $\exists \space T:V\rightarrow V'$ linear isomorphism such that $T{\rho_g}={\rho'_g}T\space \forall g\epsilon G$.
What I would think a more natural definition would be:
Possible different definition: Let $G$ be a group, $\rho : G\rightarrow GL(V)$ and $\rho' : G\rightarrow GL(V')$ be two representations of G. We say that $\rho$ and $\rho'$ are $equivalent$ (or isomorphic) if $\rho(G)$ is isomorphic to $\rho'(G)$.
Then we would get the nice property that:
Given $\rho : G \rightarrow GL_n(\mathbb{C})$, $\rho' : G \rightarrow GL_1(\mathbb{C})$, such that $\rho(g) = \text{Id}_n$ and $\rho'(g) = \text{Id}_1$ for all $g \in G$. Then we have $\rho \sim \rho'$.
Can one say if there would be problems with the possible different definition?
I feel it would be more natural if a definition of equivalent representations would only consider the images of the representations, and not the underlying vector spaces.
Firstly, see this question.
The usual notion of isomorphism implies yours: Let $\phi\colon V\to V'$ be an intertwining operator that is a linear isomorphism. Then if $\rho(g)=1$, $\rho'(g)\phi(v)=\phi(\rho(g)v)=\phi(v)$ for all $v\in V$, and $\rho'(g)=1$. The same argument the other way round shows completes the proof that $\ker\rho=\ker\rho'$, so that $\rho(G)\simeq \rho'(G)$. However, your notion is isomorphism cannot distinguish between a representation $V$ and a $V\oplus\mathbb{C}^n$ for any $n$, where $G$ acts trivially on the second factor. The major task upon meeting a new representation is to decompose it into irreducibles. For finite groups, a great too for this is character theory, which won't work for your definition.
Generally, the notion of being isomorphic for a given algebraic structure should imply that two isomorphic objects are also isomorphic as sets. Your example property shows how far your definition is from this: if you tell me you have two equivalent representations, they aren't even guaranteed to be isomorphic as vector spaces, and a representation is a vector space with extra structure.