The invariance group of $f(z)=z\overline{z}$ is $U(1)$. Here I am interested in the invariance group of $f(z)=z^2\overline{z^2}$, but more-over regarding how its generalizes to multiple complex numbers: $f(z_1,z_2,...)=z_1^2\overline{z_1^2}+z_2^2\overline{z_2^2}+...$
First, consider $z=a+bi$. It follows that:
$$ z\overline{z}=a^2+b^2 \implies (z\overline{z})^2=a^4+b^4+2a^2b^2 $$
Now, I compare it to $z^2\overline{z^2}$:
$$ \begin{align} z^2\overline{z^2}&=(a^2-b^2+2iab)\overline{(a^2-b^2+2iab)}\\ &=(a^2-b^2)^2-(2iab)^2\\ &=a^4+b^4+2a^2b^2\\ &=(z\overline{z})^2 \end{align} $$
As we can see $z^2\overline{z^2}=(z\overline{z})^2$. Consequently, I expect that the invariance group of $z^2\overline{z^2}$ is the same as that of $(z\overline{z})^2$, namely of $U(1)$.
It gets interesting when I generalize the relation to multiple complex numbers. Since
$$ (z_1\overline{z_1}+z_2\overline{z_2}+...)^2\neq z_1^2\overline{z_1^2}+z_2^2\overline{z_2^2}+... $$
Then it seems to me that the invariance group of each of the two relations differs. In the first case, we have $U(n)$, whereas as in the second case it remains as $U(1)$. Is this correct?
edit: Maybe I have a proof:
$$ \begin{align} f(z_1,z_2)&=(z_1\overline{z_1})^2+(z_2\overline{z_2})^2\\ \implies f(Uz_1,Uz_2)&=(Uz_1\overline{Uz_1})^2+(Uz_2\overline{Uz_2})^2\\ &=(Uz_1\overline{z_1}\overline{U})^2+(Uz_2\overline{z_2}\overline{U})^2\\ &=(U\overline{U})^2((z_1\overline{z_1})^2+(z_2\overline{z_2})^2)\\ \implies U\overline{U}&=1 \end{align} $$
edit2: actually, I think I realized my mistake... its all U(1) because $z$ is a 1x1 matrix. For higher U(n), z would have to be a nx1 vector.