Schurs Lemma for endomorphisms Schur's lemma states that for an irrep $(\varphi,V)$ any endomorphism $\phi: V \mapsto V$ is given by a scalar mapping.

Lets say we are in the complex case, then this would mean:
$\phi = \lambda I$

Since any homomorphism between two representations is defined by its property $\varphi_1 = \phi^{-1} \varphi_2 \phi$, this would mean:

$\varphi_1 = (\lambda I)^{-1} \varphi_2 (\lambda I) = \varphi_2$

Therefore only a single irrep over $V$ exists, since $\varphi_1 = \varphi_2$.
What am I missing?

This should be false, since I found for the group $D_4$ at least two different irreps:
$$1:  \langle \{ \begin{pmatrix} 0 &amp; 1 \\ -1 &amp; 0 \end{pmatrix}, \begin{pmatrix} 1 &amp; 0 \\ 0 &amp; -1 \end{pmatrix} \}\rangle$$
$$2:  \langle \{ \begin{pmatrix} 0 &amp; -1 \\ 1 &amp; 0 \end{pmatrix}, \begin{pmatrix} -1 &amp; 0 \\ 0 &amp; 1 \end{pmatrix} \}\rangle$$

where $\phi$ should equal $\lambda I$, but for all $\lambda$
$$\begin{pmatrix} 1 &amp; 0 \\ 0 &amp; -1 \end{pmatrix} \ne (\lambda I)^{-1} \begin{pmatrix} -1 &amp; 0 \\ 0 &amp; 1 \end{pmatrix} (\lambda I).$$

Where goes everything wrong? Thanks for any help!
In the first two sentences you state versions of Schur's lemma involving an endomorphism $\phi$ (although the first statement is not correct unless you already assume yourself to be over an algebraically closed field). Starting from the third sentence you take $\phi$ to be a morphism, not an endomorphism. But what does it mean for a morphism between two representations to be multiplication by a scalar?

The correct version of Schur's lemma for a morphism between two irreducible repesentations is that the morphism must be either zero or invertible. It is zero if they are not isomorphic and invertible if they are. Note that in your second equation you assume that $\lambda \neq 0$.

Unitary has nothing to do with any of this.
Doesn't the condition for two irreps to be equivalent require the transformation ($\phi$ in your case) to be unitary? If $|\lambda|\neq 1$, the corresonding transformation is non-unitary.