Proposition: (Schur's Lemma) Let $(\rho,V)$ and $(\tau,W)$ be irreducible representations of a finite group $G$. Then $$ \text{Hom}_G(V,W) \cong \begin{cases} \mathbb{C} & \text{ if $\rho \cong \tau$}; \\ 0 & \text{ otherwise} \end{cases}. $$ Here $\text{Hom}_G(V,W)$ is the collection of all $G$-equivariant linear maps $T:V \rightarrow W$, or equivalently all $\mathbb{C}G$-module homomoprhisms between $V$ and $W$. I understand that if $\rho \ncong \tau$ then the only $G$-equivariant map between them is trivial, its an immediate consequence of $V$ and $W$ being irreducible $\mathbb{C}G$-modules.
I'm having a conceptual issue right at the end of the proof for the latter case:
Proof: If $\rho \cong \tau$ then there is a $G$-equivariant linear map $T:V \rightarrow W$ which is an isomorphism. Therefore we can identify $\text{Hom}_G(V,W)$ with $\text{End}_G(V)$m and so it suffices to show the endomorphism ring is $1$-dimensional. Take any non-zero $\varphi \in \text{End}_G(V)$, then as a non-zero operator on a finite dimensional vector space it has some eigenvalue $\lambda$ with a non-zero eigenvector $\xi$. The map $T:V \rightarrow V$ defined by $T = \varphi - \lambda \cdot 1_V$ is an element of the endomorphism of ring, but $\ker T \neq \{0\}$ since $T(\xi) = 0$. Since $V$ is an irreducible $\mathbb{C}G$-module it follows that $\ker T = V$ therefore $\varphi = \lambda \cdot 1_V$. Therefore we have an isomorphism $\text{End}_G(V) \rightarrow \mathbb{C}$ by $\varphi \mapsto \lambda$, and in particular we can compute $\lambda$ as $\lambda = \text{Trace}(\varphi)/\dim V$.
Everything makes sense to me until the end. Because from what I understand, we picked any eigenvalue of $\varphi$ and showed that $\varphi$ acts as a scalar multiple of $\lambda$. But then how is the isomorphism between the endomorphism ring and $\mathbb{C}$ well-defined, since we could have picked a different eigenvalue in the construction, giving rise to a different scalar multiple of the identity. So I suppose I am asking how is this construction well-defined in the first place, because I don't see why in general $\varphi(v) = \lambda_1v = \lambda_2(v)$ if $\lambda_1,\lambda_2$ are two distinct eigenvalues of $\varphi$. Thanks in advance for the clarification, and let me know if I can be more precise anywhere.