What is the symmetry group of $(z, z^*) \in\mathbb{C}^2$? I'm curious about the symmetry group of $(z, z^*)\in\mathbb{C}^2$; that is, the complex vector in $\mathbb{C}^2$ consisting of a complex number and it's conjugate. 
The first thing I though of was to explicitly construct the symmetry matrices that left this quantity invariant. So for $a,b,c,d\in\mathbb{C}$, we want to try and satisfy the condition
$$\left(\begin{array}{cc}a &b\\c&d\end{array}\right)\left(\begin{array}{c}z\\z^*\end{array}\right)=\left(\begin{array}{c}z\\z^*\end{array}\right)$$
such that $ad-cb\neq0$ (so that the transformation is invertible). Besides the identity matrix, one can find some particular equations that must be satisfied by the matrix, but they don't suggest any particular properties (like orthogonality or unitarity):
$$(a-1)(d-1)=bc.$$
Thinking about this more intuitively, if you're fixing a particular vector in $\mathbb{C}^2\simeq\mathbb{R}^4$ and considering transformations that leave it invariant, I think the answer might just be rotations of a 3-sphere $\mathbb{S}^3$, so the matrices above might just be representations of $SO(3)$.
Can anyone help confirm or correct either of those two approaches?
 A: I'll answer the question in your body text first, that is, what are all the complex $2 \times 2$ matrices such that 
$$\left(\begin{array}{cc}a &b\\c&d\end{array}\right)\left(\begin{array}{c}z\\z^*\end{array}\right)=\left(\begin{array}{c}z\\z^*\end{array}\right)$$
for all complex numbers $z$?
The answer is "Only the identity matrix". Notice that your matrix $A = \left(\begin{array}{cc}a &b\\c&d\end{array}\right)$ has to fix both of the complex vectors $\left(\begin{array}{c}1\\1\end{array}\right)$ and $\left(\begin{array}{c}i\\-i\end{array}\right)$, but these form a basis for $\mathbb{C}^2$, and so $A$ must in fact fix all of $\mathbb{C}^2$, i.e., it is the identity matrix. 
The fundamental reason for this is that complex conjugation is not a $\mathbb{C}$-linear map, and so you're asking for matrices that fix a non-linear subset of $\mathbb{C}^2$ given by $z_2 = z_1^*$, and so it's not surprising that only the identity works. 
However, your title question and last paragraph suggest that you're not really interested in this space as a subset of $\mathbb{C}^2$, but maybe just as a subset of $\mathbb{R}^4$, and here, it is a (real)-linear subspace! So let's recast the question.
$\mathbb{C}^2 \cong \mathbb{R}^4$ via the map $(x+iy,z+iw) \mapsto (x,y,z,w)$, and so we're looking for elements $A \in GL(4,\mathbb{R})$ that satisfy
$$A \left(\begin{array}{c}x\\y \\ x \\ -y\end{array}\right)=\left(\begin{array}{c}x\\y \\ x \\ -y\end{array}\right),$$ i.e., that fix the linear subspace $W$ given by the equations $z = x, w = -y$. 
This subspace $W$ has basis $\left\{ \left(\begin{array}{c}1\\0 \\ 1 \\ 0\end{array}\right),  \left(\begin{array}{c}0\\1 \\ 0 \\ -1\end{array}\right)\right\}$, which can be extended to an (orthogonal) basis $\left\{ \left(\begin{array}{c}1\\0 \\ 1 \\ 0\end{array}\right),  \left(\begin{array}{c}0\\1 \\ 0 \\ -1\end{array}\right), \left(\begin{array}{c}1\\0 \\ -1 \\ 0\end{array}\right), \left(\begin{array}{c}0\\1 \\ 0 \\ 1\end{array}\right)\right\}$ of $\mathbb{R}^4$. 
With respect to this basis, $A$ changes to:
$$\left(\begin{array}{cccc}1 \ 0 \ a \ b\\ 0 \ 1 \ c\ d\\ 0 \ 0 \ e\ f\\ 0 \ 0 \ g\ h\end{array}\right)$$
with $eh - fg \neq 0$. 
This is because $A$ must leave $W$ invariant and act as the identity there. If you want a name for this subgroup, it's isomorphic to the semidirect product $M(2,\mathbb{R}) \rtimes GL(2,\mathbb{R})$ under left multiplication.
Edit: I see now that you may have been asking the question for only one particular vector $(z, z^*)$ and not all vectors of this form. But then it really isn't like orthogonal or unitary matrices at all, which have to fix entire invariants of vectors (like length or angle). In any case, for any non-zero vector in $\mathbb{C}^2$, by extending it to a basis as I did above, we can see that the matrices that fix it are (with respect to that basis)
$$\left(\begin{array}{cc}1 \ b\\ 0 \ d\end{array}\right)$$
with $ d \neq 0$. 
If you want a name for this subgroup, it's isomorphic to the semidirect product $M(1,\mathbb{C}) \rtimes GL(1,\mathbb{C}) \cong \mathbb{C} \rtimes \mathbb{C} \setminus \{0\}$ under left multiplication.
The place I think you're confused is the difference between a point and a vector. If you're looking for all Eucldiean symmetries (isometries) that fix a point in $\mathbb{R}^4$, then that stabilizer is isomorphic to SO(3), regardless of the form of the point. But you asked to fix a vector, which by linearity fixes an entire line, so the answer is different.
