More understanding of Schur's Lemma (representation theory)

Question

One of the famous theorem from Schur's Lemma is the following:

Let $\vartheta$: $s \rightarrow D(s)$ be an irreducible representation of the finite group $G$ in the representation space $V$, and let $Q$ be a linear operator $V\rightarrow V$ habing the symmetry of $\vartheta$; i.e., $QD(s) = D(s)Q$ for all $s\in G$. Then $Q = \lambda\cdot I$, where $I$ is the identity matrix. This $\lambda$ is the eigenvalue of $Q$.

My question is:

1. Does this theorem say that if any square matrix $Q$ satisfying the above requirements, then this $Q$ can be written as $\lambda\cdot I$, for any eigenvalue $\lambda$?

2. Is $Q = \lambda\cdot I$ based on a special basis? (I think this is like a basis transformation or similar transformation). If it is, how to find such transformation?

thanks

Answer 1

1) This theorem says that, for any square matrix $Q$ such that $QD(s) = D(s)Q$ for all $s \in G$, there exists a scalar $\lambda$ such that $Q = \lambda . I$.

2) If $Q = \lambda . I$ holds in some basis, then it holds in all bases. No "special" choice of basis is required.