Let $B$ be a quaternion algebra over a field $\mathbb{F}$ with generators $i,j\in B$ with $i^2=a\in\mathbb{F}^{\times}$ and $j^2=b\in\mathbb{F}^{\times}$. Let now $i'\in B\setminus\mathbb{F}$ such that $i'^{2}=a'\in\mathbb{F}^{\times}$. I want to find $b'\in\mathbb{F}^{\times}$ such that $B\overset{\phi}{\cong}(a',b')_{\mathbb{F}}$, under which isomorphism $\phi$, $i$ maps to $i'$.
I have no idea where to start and how to find such $b'$. The only thing I have thought of is to embed $B$ into the matrix algebra $M_{2}(\mathbb{F}(\sqrt{a}))$ under the map (defined on the generators) $$i,j\mapsto \begin{pmatrix} \sqrt{a} & 0 \\ 0 & -\sqrt{a}\end{pmatrix}, \begin{pmatrix} 0 & b \\ 1 & 0\end{pmatrix}$$ and then take the $\mathbb{F}$-algebra $\mathbb{F}[i']=\mathbb{F}\oplus\mathbb{F}i'$, generated by $i'$. Then, we have an embedding $\mathbb{F}[i']\hookrightarrow B\hookrightarrow M_{2}(\mathbb{F}(\sqrt{a}))$ and we want to find some appropriate $j'$ to take the quaternion algebra $\mathbb{F}[i']+j'\mathbb{F}[i']$. But...
(*) Here $(a',b')_{\mathbb{F}}$ denotes the quaternion algebra over $\mathbb{F}$ with generators $i',j'$ with $i'^2=a'\in\mathbb{F}^{\times}$ and $j'^2=b'\in\mathbb{F}^{\times}$.
Skolem Noether theroem tells that you can extend the non trivial automorphism of $F(i')/F$ to an inner automorphism $Int(j')$ of $B$.
So in particular, it means that $i'j'=-j'i'$. It shouldn't be difficult to show that $1,i',j',i'j'$ is a basis (show that they are linearly independent)
Note that the restriction of this automorphism to $F(i')$ has order $2$, so ${j'}^2$ commutes with $i'$. Since it commutes with $j'$, it commutes any element of $B$, hence lies in the center $F$.