Homomorphisms from $\mathbb{C}$ to $M_2(\mathbb{R})$ are conjugate 
Let $\phi_1$ and $\phi_2$ be two ring homomorphisms from $\mathbb{C}$ to $M_2(\mathbb{R})$. Show that there exists $g\in GL_2(\mathbb{R})$ such that $\phi_2(x) = g\phi_1(x)g^{-1}$ for all $x\in\mathbb{C}$.

$\phi(1) = I$ by definition of ring homomorphism.  By additivity, $\phi(-1) = -I$.  $\phi(i)^2 = \phi(i^2) = \phi(-1) = -I$ and hence the minimal polynomial of $\phi(i)$ in field $\mathbb{R}$ is $x^2+1$.  If we consider the rational forms of $\phi(i)$, then we see that there exists $g\in GL_2(\mathbb{R})$ such that $\phi_2(i) = g\phi_1(i)g^{-1}$.
If we can show that $\phi(r) = rI$ for all $r\in\mathbb{R}$, then we are done because $1$ and $i$ are a basis for $\mathbb{C}$ over $\mathbb{R}$.  It's easy to see that $\phi(q) = q$ for all $q\in\mathbb{Q}$. I want to use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$, but I'm stuck here.
Any suggestion? Thanks.
 A: Unfortunately, I find this problem is wrong because of the following reasons:
Suppose this statement is true. Then for each such homomorphism $\phi$, there exists $g\in GL_2(\mathbb{R})$ such that $\phi(r) = g\psi(r)g^{-1} = rI$. 
Lemma 1. Diagonalization of skew-symmetric matrices in $\mathbb{R}^{2\times 2}$
$$P^{-1}\begin{pmatrix}a & b\\-b & a\end{pmatrix}P = \begin{pmatrix}a+bi&\\&a-bi\end{pmatrix},$$ where $P = \begin{pmatrix}1 & i\\i & 1\end{pmatrix}$.
Lemma 2.
Let $E$ be an algebraic extension of $F$ and $E$ is algebraically closed. Given an automorphism $\phi$ of $F$, then there exists an automorphism $\psi$ of $E$ such that $\psi(f) = \phi(f)$ for all $f\in F$.
Lemma 3.
If $F$ is a subfield of $\mathbb{C}$ and $\phi$ is an automorphism of $F$, then there exists an automorphism $\psi$ of $\mathbb{C}$ such that $\psi(f) = \phi(f)$ for all $f\in F$.
Remark.
There are infinitely many automorphisms of field $\mathbb{C}$. That is, $|Aut(\mathbb{C})| = |Aut(\mathbb{C}/\mathbb{Q})| = \infty$.
Given an automorphism $f$ of $\mathbb{C}$ with $f(\sqrt{2}) = -\sqrt{2}$ (it exists by Lemma 2), we define $\phi(z) = P\begin{pmatrix}f(z)&\\&\overline{f(z)}\end{pmatrix}P^{-1}$. Such $\phi$ is an injective homomorphism from $\mathbb{C}$ to $M_2(\mathbb{R})$ and $\phi(\sqrt{2}) = -\sqrt{2}I$. Contradiction!
Thanks to Jiachen Huang and Ningzhe Gaoan for providing me with important insights.
