consider the following property (invariant) for complex $C^{*}$ algebras:
"$T(x)=x^{*}$ is the only non zero $\mathbb{R}$-linear map on $A$ which satisfies $T(x)T(y)=T(yx)$."
Questions:
1)Some examples?
2)Does this property implies that $A$ has trivial center?
On
Since the question does not consider $A$ represented, it seems reasonable to assume that $T:A\to A$. In that case, Tom's argument does not work as stated. But it can be saved: using his same idea, we would deduce that $A$ commutes with $U$ for each unitary in $A$; as $A$ is spanned by its unitaries, it follows that $A$ is abelian.
So all representations of $A$ are one-dimensional, and they all satisfy $\pi(xy)=\pi(y)\pi(x)$. By the uniqueness, there is a single one-dimensional representation, i.e. $A=\mathbb C$.
Suppose (as in the current version of the question) that conjugation is the only $\mathbb R$-linear map satisfying $T(xy)=T(y)T(x)$. We will see that the $C^*$-algebra $A$ must be isomorphic to $\mathbb C$.
Let the $C^*$-algebra $A$ be faithfully represented on a Hilbert space $H$. Let $U$ be any unitary on $H$. Then $$T_U(x) = U^*x^*U$$ will be an anti-automorphism of $A$. So if conjugation is the only anti-automorphism of $A$, then we require that $A$ commutes with $U$ for every unitary $U$ on the Hilbert space $H$. As every operator in $B(H)$ is a linear combination of unitaries, it follows that $A'= B(H)$. Thus $A \subseteq A'' \subseteq B(H)'= \mathbb C \cdot id_H$. Here $A'$ denotes the commutant of $A$, the set of all bounded linear operators that commute with $A$.