Let $G$ denote a finite group, and let $V$ be a complex and finite dimensional $G$ module; that is, there is a group representation $G\to GL(V)$. Finally, let $T\colon V\to V$ be an intertwining map, that is, a linear map such that $T(gv)=g T(v)$ for all $v\in V$ and $g\in G$.
By the theorem of Maschke, there are irreducible $G$-modules $V_1, V_2, \ldots, V_n$ such that $$ V=V_1\oplus V_2 \oplus \ldots \oplus V_n.$$
Question. Is it true that $$T v_j=\lambda_j v_j, \qquad \forall v_j\in V_j,$$ for some $\lambda_1,\ldots,\lambda_n\in\mathbb C$?
The lemma of Schur says that this is the case provided that $$\tag{*}T(V_j)\subset V_j.$$ This is because, in this case, $T$ restricts to an intertwining self-map of $V_j$, which is irreducible. Therefore such restriction must be a scalar multiple of the identity.
But is it true that (*) always holds with the given assumptions?
EDIT. This question arises from the following observations. Suppose that $G$ is finite and abelian, and let $L^2(G)$ denote the space of all complex-valued functions on $G$, which is a $G$-module with the representation $gf(x):=f(x-g)$. (This seemingly complicated notation hints at more general cases, with infinite groups).
Let $\chi\in L^2(G)$ denote a character, that is, a homomorphism of $G$ into $\mathbb C^\times$. Then an intertwining map $T\colon L^2(G)\to L^2(G)$ satisfies $$T\chi=\lambda_\chi \chi, $$ as it is easy to prove. And since the irreps are in this case the 1-dimensional subspaces $$ \operatorname*{span}(\chi), $$ it follows that intertwining maps of the $G$-module $L^2(G)$ are diagonalized by irreps. (As the accepted answer clearly shows, intertwining maps of other $G$-modules need not be even diagonalizable! This is the reason why I found that answer surprising and enlightening).
Let us consider an infinite and non-abelian case. Suppose that $T\colon L^2(\mathbb S^{d-1})\to L^2(\mathbb S^{d-1})$ is rotation-invariant; $$ (Tf)(R^{-1}x)=T(f(R^{-1}\cdot))(x).$$ Then $T$ is diagonalized by spherical harmonics. Precisely, letting $$\{Y_{n, j}\ :\ j=1, \ldots, N(n)\}$$ denote a complete orthonormal system of spherical harmonics of degree $n$, we have that $$ Tf=\sum_{n=0}^\infty \lambda_n \sum_{j=1}^{N(n)} \hat{f}(n, j) Y_{n, j}, $$ where we have let $\hat{f}(n, j)$ denote the coefficient $\langle f | Y_{n, j}\rangle.$
This latter example actually is a consequence of the lemma of Schur. Indeed, the decomposition of the $SO(d)$-module $L^2(\mathbb S^{d-1})$ into irreps is precisely $$ \bigoplus_{n=0}^\infty \operatorname*{span}\{ Y_{n,j}\ :\ j=1, \ldots, N(n)\}, $$ and these irreps are pairwise non-isomorphic, because $N(n)$ is not a constant. (Actually, $N(n)$ has a well-known combinatorial expression, which there is no need to write down explicitly here).
It is not necessarily true that $T(V_j) \subset V_j$ for all $j$. For instance, consider the representation of $G = \Bbb Z/n \Bbb Z$ given by $$ \rho(k) = \pmatrix{\omega^k & 0\\0 & \omega^k}, $$ where $\omega$ is the $n$th root of unity $\omega = e^{2 \pi i / n}$. We note that in this case, $$ T = \pmatrix{0&1\\1&0} $$ is an intertwining map (in fact, so is any $T : \Bbb C^2 \to \Bbb C^2$).