Writing $\rho(A)X=A_1XA_1^{-1}$, we have
$$
\rho(A)\rho(B)X=A_1B_1X B_1^{-1}A_1^{-1}=(AB)_1X(AB)_1^{-1}=\rho(AB)X,
$$
and $\rho(I)=I$, so this <conjugation representation> is indeed a <group representation>.

The underlying three-dimensional module is
$$
W=\mathbb C^3\cong V_1\oplus V_0.
$$
Using $M_3(\mathbb C)=\operatorname{End}(W)\cong W\otimes W^*$, the <self-duality of finite-dimensional SU2 representations>, and $V_1\otimes V_1\cong V_2\oplus V_0$, we obtain
$$
\begin{aligned}
M_3(\mathbb C)
&\cong(V_1\oplus V_0)\otimes(V_1\oplus V_0)\\
&\cong(V_2\oplus V_0)\oplus V_1\oplus V_1\oplus V_0.
\end{aligned}
$$
Therefore the irreducible decomposition is
$$
\boxed{M_3(\mathbb C)\cong V_2\oplus2V_1\oplus2V_0}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 19I / b / Solution

If $G$ acts on a vector space $W$, it acts on $\operatorname{End}(W)$ by conjugation:
$$
g\cdot X=\rho(g)X\rho(g)^{-1}.
$$
Under $\operatorname{End}(W)\cong W\otimes W^*$, this is the <tensor product of group representations> $W\otimes W^*$.

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 Incoming links: Conjugation representation