Let $A$ be a $*$-algebra and $\phi: A \mapsto A$ be a map such that $\phi(x)^*=\phi(x^*)$ and $\phi\circ \phi=id$. Is it then true that $\phi$ will be an antihomomorphism? Do I need more assumptions on $\phi$?
In fact my $\phi$ is of the form $\phi=T\psi T^{-1}$ where $T: A \to A$ is an isomorphism of underlying vector space of $A$ and $\psi$ is a $*$ preserving antihomomorphism and is involutive. Will then $\phi$ be an antihomomorphism?
No, it is not. For example, if $p\in M_n(\mathbb C)$ is a nontrivial projection, then $$ \phi(a)=a-2pap $$ is a counter example.
Edit: Another counter example is the identity map on any noncommutative star algebra.