A consequence of Schur's lemma for two equivalent irreducible representations of a finite group $G$ on a complex vector space $V$, $r^1(G,V)$ and $r^2(G,V$), is that linear $G$-morphisms between $r^1$ and $r^2$ must be of the form $cI$ where c is a complex scalar.
But what goes wrong if we try to use a "change of coordinate" matrix as the $G$-morphism?
As an explicit example, consider any two-dimensional irreducible representation. Then consider a standard rotation matrix $T_r$ in terms of some angle $\theta$. Through conjugation, I can map the matrix representation of any element of $G$ in $r^1$ to another matrix, and let that be the matrix representation of the same group element for $r^2$ (I'm constructing $r^2$ this way). So by construction we have $r^2 T_r = T_r r^1$.
And I would think that this same matrix $T_r$ could be used for all group elements, because conceptually it amounts to redrawing the coordinate axes, which shouldn't affect the way group actions behave.
But clearly there's a mistake somewhere in my reasoning here. An explicit counter-example could be very helpful.
Edited addition:
I'm reframing my question by using the explicit example of $G = S_3$ and $V = \mathbb{C}^2$.
I may define $\rho^1$ as
$e \mapsto \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$, $(123) \mapsto\begin{bmatrix} -1/2 & -\sqrt{3}/2\\ \sqrt{3}/2 & -1/2\end{bmatrix}$, $(132) \mapsto \begin{bmatrix} -1/2 & \sqrt{3}/2\\ -\sqrt{3}/2 & -1/2\end{bmatrix}$,
$(12) \mapsto \begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}$, $(13) \mapsto \begin{bmatrix} -1/2 & \sqrt{3}/2\\ \sqrt{3}/2 & 1/2\end{bmatrix}$, $(23) \mapsto \begin{bmatrix} -1/2 & -\sqrt{3}/2\\ -\sqrt{3}/2 & 1/2\end{bmatrix}$
Then I define
$f = \begin{bmatrix}\cos{\theta} & \sin{\theta}\\ -\sin{\theta} & \cos{\theta}\end{bmatrix}$
for any real value of $\theta$, and define $\rho^2$ as
$$ \rho^2(g)= f \, \rho^1(g) \, f^{-1}, \qquad g \in S^3.$$
I've checked that $\rho^2$ is still a representation of $S^3$ , i.e. it obeys the group element multiplication table for $S^3$. I also know $\rho^1$ and $\rho^2$ are irreducible. So, it seems to me I am in conflict with part (2) of Schur's lemma, which I copy below from the textbook by Serre:
Let $\rho^1: G \to \mathbf{GL}(V_1)$ and $\rho^2: G \to
> \mathbf{GL}(V_2)$ be two irreducible representations of $G$, and let
$f$ be a linear mapping of $V_1$ into $V_2$ such that $\rho_s^2 \circ$
$f$ = $f \circ \rho_s^1$ for all $s \in G$. Then:(1) If $\rho_1$ and $\rho_2$ are not isomorphic, we have $f = 0$
(2) If $V_1 =V_2$ and $\rho^1 = \rho^2$, $f$ is a homothety (i.e., a
scalar multiple of the identity).
How is there no conflict with (2), given the example I posed above?
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