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Just for interest, I point out that it is well-known that for every finite group $G,$ we can make the group algebra $\mathbb{C}G$ into a $G \times G$-module, by setting $a(g,h) = g^{-1}ah$ for all $a \in \mathbb{C}G,$ for all $(g,h) \in G \times G.$ It is easy to check that this representation is multiplicity-free. In fact, calculating the trace with respect to the natural group basis for $\mathbb{C}G,$ we see that ${\rm trace}(g,h) = |C_{G}(g)|$ if $g$ and $h$ are conjugate in $G,$ and $0$ if they are not conjugate. Hence the character of $G \times G$ afforded by this representation is $\sum_{i=1}^{k} \overline{\chi}_{i} \otimes \chi_{i}$, where $\{ \chi_{i} : 1 \leq i \leq k \}$ are the irreducible charactes of $G.$ It is also easy to see directly that the algebra of $G \times G$ endomorphisms of this module is isomorphic to the commutative ring $Z(\mathbb{C}G).$ This way of looking at the group algebra as a $G \times G$-module ( or as a bimodule for $G$) was profitably exploited in modular representation theory by J.A. Green. But in the context of this question, it illustrates that the existence of a faithful multiplicity free representation puts very little restriction on the structure of the group. Notice that if we take the group basis for $G,$ the module may be viewed as a transitive permutation module, and that the stabilizer of the identity element $1_{G}$ is the "diagonal" subgroup $\Delta(G) = \{ (g,g) : g \in G \}.$ It is easy to check that the module is faithful if and only if $G$ has trivial center.