Let $ \pi: G \to GL_d(\mathbb{C}) $ be a degree $ d $ irrep of a finite group $ G $ with character $ \chi_\pi $. Consider the linear map $ T: \mathfrak{gl}_d(\mathbb{C}) \to \mathfrak{gl}_d(\mathbb{C}) $ given by $$ T: M \mapsto \frac{d}{|G|}\sum_{g \in G} \chi_{\pi}(g) \, \pi(g) \, M \, \pi(g^{-1}) $$ Does $ T $ always have a nontrivial kernel? For $ d=1 $ the answer is yes (unless $ G $ is the trivial group). Indeed in this case $ T $ is identically zero since the sum of all $ n $th roots of unity is $ 0 $. (we have to exclude the case of the trivial group since in case the sum of roots of unity is 1 and $ T=1 $ is the identity.
My main question is: does $ T $ always have nontrivial kernel when $ \pi $ is irreducible? And I'm interested generally in knowing about the kernel of $ T $.
Edit: I've gone back and slightly edited the definition of $ T $ to clarify notation. I've also added the standard scalar factor in front. As pointed out in the answer below, for a scalar matrix $ M $ then $ T(M)=\Big( \sum_{g\in G} \chi_\pi(g) \Big) M $ which vanishes by character orthogonality of $ \chi_\pi $ with the trivial character (as long as $ \chi_\pi $ is not itself trivial). Thus, as long as $ \pi $ is not the trivial irrep, then $ ker(T) $ is at least dimension $ 1 $ since it contains all the scalar matrices. If $ \pi $ is trivial then $ T $ is just the identity operator.
Scalar matrices are in the kernel, so it's always nontrivial (assuming $\pi$ is nontrivial).
Indeed, $T$ is an application of the isotypical projector for the dual rep $\pi^\ast$ to the rep $\mathrm{End}(V)\cong V\otimes V^\ast$, where $V=\mathbb{C}^d$ in your post. Thus, the image is the $\pi^\ast$-isotypical component of this rep. This component is comprised of $V^\ast$s with multiplicity that can be computed using the standard inner product for characters. So we can compute
$$ \begin{array}{ccl} \dim\mathrm{img}\,T & = & (\dim V^\ast)\big(\,\overbrace{\dim\hom_G(V^\ast,V\otimes V^\ast}^{\textrm{multiplicity of }V^\ast\textrm{ in }V\otimes V^\ast})\big) \\ & = & \frac{d}{|G|}\sum_{g\in G}\chi(g)|\chi(g)|^2, \end{array}$$
and from that compute $\dim\ker T$ by subtracting from $|\mathrm{End}(V)|=d^2$ (by rank-nullity).
Some more detail.
We can interpret elements of $V=\mathbb{C}^d$ as column vectors, and of $V^\ast$ as row vectors, and then $G$ acts on $V$ via left-multiplication by $\pi(g)$, and on $V^\ast$ via right-multiplication by $\pi(g)^{-1}$ (the so-called "contragredient" action). The iso $V\otimes V^\ast\cong\mathrm{End}(V)$ given by $u\otimes v^T\mapsto uv^T$ (i.e., the abstract tensor product effectively becomes the Kronecker product) is thus an equivalence of reps, where $G$ acts on $\mathrm{End}(V)$ by conjugation, $g\cdot M:=\pi(g)M\pi(g)^{-1}$.
Write $E=\mathrm{End}(V)$, so $\tau:G\to\mathrm{GL}(E)$ is defined by $\tau(g)(M)=\pi(g)M\pi(g)^{-1}$. We can extend linearly to an algebra rep $\tau:\mathbb{C}[G]\to\mathrm{End}(E)$, and then rewrite $T(M)$ in terms of the isotypical projector $P_{\pi^\ast}=\frac{d}{|G|}\sum \chi_{\pi^\ast}(g^{-1})g\in\mathbb{C}[G]$ as
$$ \begin{array}{ccl} T(M) & = & \frac{d}{|G|}\sum \chi_\pi(g)\pi(g)M\pi(g)^{-1} \\ & = & \frac{d}{|G|}\sum \chi_{\pi^\ast}(g^{-1})\tau(g)(M) \\ & = & \tau\Big(\frac{d}{|G|}\sum \chi_{\pi^\ast}(g^{-1})g\Big)(M) \\ & = & \tau(P_{\pi^\ast})(M). \end{array} $$