$\langle M_2(\mathbb{R}), \cdot \rangle$ with $\langle \mathbb{R}, \cdot \rangle$ where $\phi(a)$ is the determinant of matrix $M_{2\times 2}$.
I don't think $\phi$ is an isomorphism because I think it is not onto. For a real number $r \in \mathbb{R}$, $\phi(r) \neq M_{2\times 2}$. In other words, we cannot use the inverse of $\phi$, a determinant $a$, and get a $M_{2\times 2}$.
If the two binary spaces were $\langle M_1(\mathbb{R}), \cdot \rangle$ with $\langle \mathbb{R}, \cdot \rangle$, then I think $\phi$ would have passed the onto check because $a(\mathbb{R}) = M_{1\times 1}$.
Is my reasoning correct?
Edit with additional work/clarifications:
The map is not injective because the transpose of matrix $a$ will have the same determinant as $a^T$. Therefore $\phi$ is not an isomorphism.
Best Answer
The map is surjective. For $r\in\Bbb{R}$ the expression $\phi(r)$ makes no sense, because the domain of $\phi$ is $M_2(\Bbb{R})$, not $\Bbb{R}$.
In stead, show that the map is not injective; show that there are two matrices with the same determinant.