Prove that the rings $\mathbb{C}((z;\sigma))$ and $\mathbb{H}((x))$ are not isomorphic.

Question

Prove that the division rings $\mathbb{C}((z;\sigma))$ and $\mathbb{H}((x))$ are not isomorphic. The following statement is needed: \begin{equation} \text{any ring automorphism $\phi: \mathbb{R}((x)) \rightarrow \mathbb{R}((x))$ has the form:} \end{equation} \begin{equation} \forall a \in \mathbb{R}, \phi(a)=a; \end{equation} \begin{equation} \exists a_i \in \mathbb{R} \text{ such that } a_1 \neq 0 \text{ and } \phi(x) = a_1 x + a_2 x^2 + ... \end{equation} Results to help with this question:

• $\nexists a \in \mathbb{H}((x))$ such that $a^2=a_1 x + a_2 x^2 + ...$ and $a_1 \neq 0$. Therefore it does not exist in $\mathbb{R}((x)) \subseteq \mathbb{H}((x))$.
• The centre of $\mathbb{H}((x))$ is $\mathbb{R}((x))$ and the centre of $\mathbb{C}((z;\sigma))$ is $\mathbb{R}((x^2))$, and therefore if there exists an isomorfism, it takes the centre of a ring to the centre of the other one.

Edit: to clarify, both $\mathbb{C}((z;\sigma))$ and $\mathbb{H}((x))$ are the rings of Laurent series with elements of the form $\sum_{n=m}^\infty a_n x^n$ with coefficients $a_i$ in either $\mathbb{C}$ or $\mathbb{H}$, the quaternion ring. The ring $\mathbb{C}((z;\sigma))$ is like $\mathbb{C}((z))$ but with the multiplication of elements defined the following way: $xb=\sigma(b)x$ and extending it to $\sigma^n(b)$: \begin{equation} \sigma^n(b)= \begin{cases} b & \text{if $n$ is even} \\ \overline{b} & \text{if $n$ is odd} \end{cases} \end{equation}

I have had 3 main attempts at this, all reducing the case to $\mathbb{R}((x))$ since it's contained in both rings:

1. Trying to get an element where studying $\phi(a^2)=\phi(a)\phi(a)$ wouldn't work, but without any luck.
2. I have tried working with the centres of both rings, which are $Z(\mathbb{H}((x)))=\mathbb{R}((x))$ and $Z(\mathbb{C}((x;\phi)))=\mathbb{R}((x^2))$ and seeing that for a $t=a_0 + a_1 x + a_2 x^2 + ...$, a $\phi$ described as above didn't take it to $\mathbb{R}((x^2))$ but it didn't seem to be enough because ultimately I am taking the definition of an automorphism and applying it to $t$ (?).
3. Assuming $\phi$ the isomorphism between the two rings exists, then there exists one between its centres: $\phi:\mathbb{R}((x)) \rightarrow \mathbb{R}((z^2))$ that takes $x \rightarrow z^2$. With this we can create a ring automorphism $\gamma:\mathbb{R}((x)) \rightarrow \mathbb{R}((x))$ the following way: $\gamma = \psi \circ \phi^{-1}$ so we know $\gamma$ is as described above. And I know I need to reach a contradiction with the fact stated in my first attempt, but can't seem to find an element $a^2$ and finalise it.

I would appreciate any type of suggestion as to how to proceed from here.