The following is an abstract (Banach) algebraic take on Werner's construction. Let $A=\ell^1(\mathbb Z)$ with convolution (but in general $A$ is any Banach algebra). We turn the dual space $A^*$ into an $A$-bimodule (though in our example, $A$ is commutative) by dualising the actions:
$$ (a\cdot f)(b) = f(ba), \quad (f\cdot a)(b) = f(ab) \qquad
(f\in A^*, a,b\in A). $$
Then you can check that $T:A^*\rightarrow A^*$ is $\mathbb Z$-equivariant if and only if $T$ is a module homomorphism: I'll write $T\in\hom_A(A^*)$ to mean $T(f\cdot a) = T(f)\cdot a$ (the right action is more appropriate in the non-commutative group setting). If $e\in A$ is the unit then
$$ T(f)(a) = T(f)(ae) = (T(f)\cdot a)(e) = T(f\cdot a)(e)
\qquad (T\in \hom_A(A^*), f\in A^*, a\in A). $$
So again $T$ is determined uniquely by $m\in A^{**}, m(f) = T(f)(e)$. Conversely, given $m\in A^{**}$ we can define $T\in\hom_A(A^*)$ by this relation. So $\hom_A(A^*)\cong A^{**}$. A similar result holds if $A$ only has a bounded approximate identity.

So $A^{**}$ becomes an algebra for the product induced from $\hom_A(A^*)$. Define $$ (m\cdot f)(a) = m(f\cdot a) \qquad (m\in A^{**}, f\in A^*,a\in A).$$
Then if $m_i$ is associated to $T_i$ for $i=1,2$,
$$ (m_1m_2)(f) = (T_1\circ T_2)(f)(e) = T_1(T_2(f))(e)
= m_1(T_2(f)). $$
However, for $a\in A$,
$$ T_2(f)(a) = m_2(f\cdot a) = (m_2\cdot f)(a) $$
and so
$$ (m_1m_2)(f) = m_1(m_2\cdot f) $$
This is precisely the "1st Arens product", say $\Box$, as studied in Banach algebra theory. By making the symmetric choices we get the "2nd Arens product", say $\diamond$, which might differ. If $A$ is commutative then $m_1\Box m_2 = m_2\diamond m_1$ and so $\hom_A(A^*)$ is commutative precisely when the Arens products agree.

However, Young showed in "The irregularity of multiplication in group algebras"
Quart J. Math. Oxford Ser. (2) 24 (1973), 59–62, MathSciNet that for any locally compact group $G$, we have that $L^1(G)$ is not Arens regular. So Werner's question has a negative answer for all Abelian groups (with the axiom of choice!)