A mapping $\displaystyle *$ from $\displaystyle F_{n}$ to $\displaystyle F_{n}$ is called adjoint if
$\displaystyle \begin{array}{{>{\displaystyle}l}} ( A+B)^{*} =A^{*} +B^{*}\\ ( AB)^{*} =B^{*} A^{*}\\ \left( A^{*}\right)^{*} =A;\ \ \ \ \ \ \ \ \end{array}$for all $\displaystyle A,B\in F_{n}$
If we know $\displaystyle \psi $ is an automorphism of $\displaystyle F_{n}$ in such a way that $\displaystyle \psi ( A+B) =\psi ( A) +\psi ( B)$ and $\displaystyle \psi ( AB) =\psi ( A) \psi ( B)$ such that $\displaystyle \psi ( \lambda ) =\lambda $ for every scalar matrix $\displaystyle \lambda \in F_{n}$. then there is an element $\displaystyle P\in F_{n}$ such that $\displaystyle \psi ( A) =PAP^{-1}$ for all $\displaystyle A\in F_{n}$.
From this theorem, prove that
if $\displaystyle \ast $ is an adjoint of $\displaystyle F_{n}$ such that $\displaystyle \lambda ^{*} =\lambda $ for every scalar matrix $\displaystyle \lambda $ then there exists a matrix $\displaystyle P\in F_{n}$ such that $\displaystyle A^{*} =PA'P^{-1}$ for every $\displaystyle A\in F_{n}$.
$\displaystyle F_{n}$ is the set of all $\displaystyle n\times n$ matrices over $\displaystyle F$.
Please give me hints to prove this.
Edit: Moreover prove that $P^{-1}P'$ must be scalar.