are invertible matrices in an inner $^*$-automorphism on $M_n$ necessarily unitary? If I have a $^*$-isomorphism $\varphi\colon$$M_n(\mathbb{C})$$\to$$M_n(\mathbb{C})$ given by $x\mapsto{yxy^{-1}}$ for some $y\in$$GL_n$ (so this is an inner $^*$-automorphism -- which is basically an inner automorphism that is also a $^*$-homomorphism given by $y\in$$GL_n$).
Then is it true that $y^*=y^{-1}$? (in other words, is it also true that $y\in$$U_n$?)
 A: Note that for all $x$, we have
$$
(yxy^{-1})^{*} = yx^*y^{-1} \implies\\
y^{-*}x^*y^{*} = yx^*y^{-1} \implies\\
x^*[y^*y] = [y^*y]x^*.
$$
That is, $y$ is such that $y^*y$ commutes with all elements of $M_n$ (noting that $x \mapsto x^*$ is bijective). This necessarily implies that $y^*y$ is a multiple of the identity. Because $y$ is invertible, it must be a non-zero multiple of the identity. What conclusion we reach from here depends on the specifics of the ring from which the entries of the matrices are taken.
In the case of real matrices with the transpose and complex matrices with the conjugate-transpose, we can conclude that $y^*y$ must be a positive-definite matrix, which means that for $y^*y$ to be a multiple of the identity we must have $y^*y = tI$ for some $t > 0$. If we take $z = t^{-1/2} y$, then we find that $z$ is orthogonal/unitary (in other words, $y$ is "orthogonal/unitary up to scaling") and that the map $x \mapsto zxz^{-1}$ is precisely the same as the map $x \mapsto yxy^{-1}$. So, $y$ is not necessarily orthogonal/unitary, but there is an orthogonal/unitary matrix that produces the same map.
On the other hand, if we take $M_n$ to be the matrices whose entries are rational, then there is not always an orthogonal element of $M_n$ that results in the same map as $y$. For example, if we take
$$
y = \pmatrix{1&1\\1&-1},
$$
Then $x \mapsto yxy^{-1}$ is a $*$-isomorphism and there is no matrix $z \in M_n$ satisfying $z^{-1} = z^*$ for which $x \mapsto zxz^{-1}$ results in the same map.
As another example, if the entries of $M_n$ are taken from $\Bbb F_2$, then the fact that $y^*y$ is a non-zero multiple of the identity implies that $y^*y = 1 \cdot I$, which is to say that $y^* = y^{-1}$. So, not only is $y$ orthogonal up to scaling, it is actually orthogonal.
