Let $F$ be the center of the division ring $D$. Then $D$ represents its class in the Brauer group $Br(F)$. The opposite ring $D^{opp}$ represents the inverse element. The reason why for example the quaternions are isomorphic to their opposite algebra is that the quaternions are an element of order 2 in $Br(\mathbf{R})$, and hence equal to its own inverse in the Brauer group.
To get a division algebra that is not isomorphic to its opposite algebra we can use an element of order 3 in the Brauer group. One method for constructing those is to start with a (cyclic) Galois extension of number fields $E/F$ such that $[E:F]=3$. Let $\sigma\in Gal(E/F)$ be the generator. Let $\gamma\in F$ be an element that cannot be written in the form $\gamma=N(x)$, where $N:E\rightarrow F, x\mapsto x\sigma(x)\sigma^2(x)$ is the relative norm map. Consider the set of matrices
$$
\mathcal{A}(E,F,\sigma,\gamma)=\left\{
\left(\begin{array}{rrr}
x_0&\sigma(x_2)&\sigma^2(x_1)\\
\gamma x_1&\sigma(x_0)&\sigma^2(x_2)\\
\gamma x_2&\gamma\sigma(x_1)&\sigma^2(x_0)
\end{array}\right)\mid x_0,x_1,x_2\in E\right\}.
$$
A theorem of A. Albert tells us that this forms a division algebra with center $F$, and its order in $Br(F)$ is 3, so it will not be isomorphic to its opposite algebra. The theory is described for example in ch. 8 of Jacobson's Basic Algebra II. The buzzword 'cyclic division algebra' should give you some hits.
For a concrete example consider the following. Let $F=\mathbf{Q}(\sqrt{-3})$ and let
$E=F(\zeta_9)$, with $\zeta_9=e^{2\pi i/9}$. Then $E/F$ is a cubic extension of cyclotomic fields, $\sigma:\zeta_9\mapsto\zeta_9^4$. I claim that the element $2$ does not belong the image of the norm map. This follows from the fact 2 is totally inert in the extension tower $E/F/\mathbf{Q}$. Basically because $GF(2^6)$ is the smallest finite field of characteristic 2 that contains a primitive ninth root of unity. Now, if $2=N(x)$ for some $x\in E$, then 2 must appear as a factor (with a positive coefficient) in the fractional ideal generated by $x$. But the norm map then multiplies that coefficient by 3, and as there were no other primes above 2, we cannot cancel that. Sorry, if this is too sketchy.
Anyway (see Jacobson again), the product of the $\gamma$ elements modulo $N(E^*)$ is the operation in the Brauer group $Br(E/F)\le Br(F).$ Therefore the opposite algebra should correspond to the choice $\gamma=1/2$, (or to the choice $\gamma=4$, as $2\cdot4=N(2)$). So
$$
\mathcal{A}(E,F,\sigma,2)^{opp}\cong
\mathcal{A}(E,F,\sigma,1/2).
$$
and the choices $\gamma=2$ and $\gamma=1/2$ yield non-isomorphic division algebras, as their ratio $=4$ is not in the image of the norm map.
======================================================
Edit (added more details here, because another answer links to this answer): In general the cyclic division algebra construction works much the same for any cyclic extension $E/F$. When $[E:F]=n$ we get a set of $n\times n$ matrices with entries in $E$. The number $\gamma$ appears in the lower diagonal part of the matrix. The condition for this to be a division algebra is that $\gamma^k$ should not be a norm for any integer $k, 0<k<n$. Obviously it suffices to check this for (maximal) proper divisors of $n$. In particular, if $n$ is a prime, then it suffices to check that $\gamma$ itself is not a norm.
Edit^2: Matt E's answer here linear algebra over a division ring vs. over a field gives a simpler cyclic division algebra of $3\times3$ matrices with entries in the real subfield of the seventh cyclotomic field.
OK, so let me denote the elements of End$(⟨\mathbb{Z},+⟩)$ by $\theta_{i}$ , where we have $\theta_{i}(1)=i$ for some $i \in \mathbb{Z}$.
Also I define the evaluation homomorphism as $\alpha(\theta_{i})=\theta_{i}(1)$.
Now we can see that $\alpha:$End$(⟨\mathbb{Z},+⟩)$ $\rightarrow$ $⟨\mathbb{Z},+,\cdot⟩$, and i need to show that in fact $\alpha$ is a ring isomorphism.
First ill show its a ring homomorphism, to do this, observe that$\alpha(\theta_{i} + \theta_{j})=(\theta_{i} + \theta_{j})(1)=\theta_{i}(1) + \theta_{j}(1)=\alpha(\theta_{i})+\alpha(\theta_{j})$.
And also that $\alpha(\theta_{i}\theta_{j})=\theta_{i}(\theta_{j}(1))=\theta_{i}(j)=ij=\alpha(\theta_{i})\alpha(\theta_{j})$, so it is a ring homomorphism.
Now i just need to show its one-to-one and onto.
one-to-one: if $\alpha(\theta_{i})=\alpha(\theta_{j})$, then $\theta_{i}(1)=\theta_{j}(1)$, and since the homomorphisms are determined by what values they send the generators to, we must have $\theta_{i}=\theta_{j}$, i.e they are the same homomorphism. Hence $\alpha$ is one-to-one.
Onto: now $\forall x \in \mathbb{Z}$, we have $\theta_{x}$ as defined above being the homomorphism sending $1$ to $x$, so $\alpha(\theta_{x})=x$, so we can see that $\alpha$ is obviously onto.
Best Answer
You are on the right track. Note that $\phi(i)=x+y\sqrt{2}$ and $$ -1=\phi(-1)=\phi(i^2)=\phi(i)^2=(x+y\sqrt{2})^2=(x^2+2y^2)+2xy\sqrt{2} $$ Can you solve this equation?