Is conjugate transpose the "unique" way to define an involution on $M_n(\mathbb C)$? 

Definition: An involution on an algebra $A$ is a conjugate-linear map $a \mapsto a^*$, such that $a^{**}=a$, and $(ab)^*=b^*a^*.$


I can't think of any other ways to define the involution on $M_n(\mathbb C)$(all $n\times n$ matrices), so I feel like it should be unique in some sense. Let 
$$f:M_n(\mathbb C) \to M_n(\mathbb C)$$
be an involution, then is $f$ necessary a conjugate transpose or having a standard form closely related to a conjugate transpose?
Thanks in advance!
 A: Yes, actually the involutions you describe are in bijective correspondance with similarity classes of non-degenerate hermitian forms on $\mathbb{C}^n$.
Precisely, let $h:\mathbb{C}^n\times \mathbb{C}^n\to \mathbb{C}$ be a non-degenerate hermitian form. Then you have the adjunction relative to $h$ : $h(u(x),y) = h(x,u^*(y))$, and $u\mapsto u^*$ is a semi-linear involution on $M_n(\mathbb{C})$.
Conversely, any involution is of this form, and two hermitian forms define the same involution iff they are similar, ie there is $\lambda\in \mathbb{C}$ such that $h=\lambda h'$.
How to see this : suppose in general you have two involutions (that is involutive anti-automorphisms) on $M_n(\mathbb{C})$, say $\sigma$ and $\tau$. They restrict to involutions on the center $\mathbb{C}$. Suppose $\sigma_{|\mathbb{C}} = \tau_{|\mathbb{C}} = \iota$. Then $\theta = \sigma\circ \tau$ is an automorphism of $\mathbb{C}$-algebra of $M_n(\mathbb{C})$ (because the actions of $\sigma$ and $\tau$ on the center cancel each other out). But it is ver standard that any automorphism of $M_n(\mathbb{C})$ is inner : so $\theta(M) = H^{-1}MH$. This means that $\sigma(M) = H^{-1}\tau(M)H$.
Conversely, given $\tau$, if $H\in GL_n(\mathbb{C})$, you can define $\sigma(M) = H^{-1}\tau(M)H$, which gives you an anti-automorphism of $M_n(\mathbb{C})$ such that $\sigma_{|\mathbb{C}} = \iota$. Now it's not hard to see that $\sigma^2 = Id$ iff $\tau(H) = \lambda H$ for some $\lambda\in \mathbb{C}$. It requires a somewhat subtle argument to see that by replacing $H$ with a scalar multiple (which doesn't change $\sigma$) you can always take $\lambda = 1$ (this follows from an easy case of the Hilbert 90 theorem).
The link with hermitian forms is the following : suppose you choose $\tau$ the usual conjugate-transpose, so in particular $\iota$ is the complex conjugation. Then the condition $\tau(H)=H$ says that $H$ is a hermitian matrix. It is the matrix of a hermitian form $h$ in the standard basis of $\mathbb{C}^n$, and $\sigma$ is the adjunction for $h$.
