Let $G$ be a countable discrete group, and $l^\infty(G) := \{f \colon G \to \mathbb{C} \mid \|f\|_\infty < \infty\}$. Let a group action of $s \in G$ on $f \in l^\infty(G)$ be defined as $(s \mathbin{.} f)(g) = f(s^{-1}g)$ for all $g ∈ G$.
Is it then true that for every $s \in G$ there exists a $u_s \in B(l^2(G))$ such that (and $l^\infty(G)\subseteq B(l^2(G))$ $$ s \mathbin{.} f = u_s f u_s^* \,? $$
I somehow have a feeling that this should be true since it is (apparently) stated in $C^∗$-Algebras and Finite-Dimensional Approximations by Brown and Ozawa (page 54).
I assume $$\ell^\infty(G)\subseteq B(\ell^2(G))$$ means the embedding $f\mapsto M_f,$ where $M_f$ stands for the multiplication operator on $\ell^2(G)$.
For $s\in G$ and $\phi\in \ell^2(G)$, we define the operator $u_s$ by $$u_s \phi = s \mathbin{.} \phi.$$ Clearly, $u_s$ is an unitary operator on $\ell^2(G)$ and $u_s^*=u_{s^{-1}}$. It is routine to check that $u_s$ does the job.
By the boundedness, it is enough to check $$u_s M_f u_s^*=M_{s\mathbin{.} f}$$ on some orthonormal basis. Let $\{e_t: t\in G\}$ be the canonic orthonormal basis of $\ell^2(G)$, easily computation shows \begin{align}u_s M_f u_s^* \,e_t&=u_sM_f\,e_{s^{-1}\,t}=u_sf(s^{-1}t)\,e_{s^{-1}\,t}\\ &=f(s^{-1}t)u_s \,e_{s^{-1}\,t}=f(s^{-1}t)\,e_t\\ &=M_{s\mathbin{.} f} \,e_t. \end{align}