So let $\Big(\dfrac{\mathit{a,b}}{\mathit{F}}\Big)$ be a quaternion Algebra over a field $F$ with char $\neq 2$, and let $i,j$ be the standard generators for the quat. Algebra, meaning $i^2=a$ and $j^2= b$. Apparently $\Big(\dfrac{\mathit{a,b}}{\mathit{F}}\Big) \simeq \Big(\dfrac{\mathit{b,a}}{\mathit{F}}\Big)$ by interchanging $i$ and $j$, $(i,j)\mapsto(j,i)$.
But a homomorphism $\phi$ of $F$-Algebras restricts to the identity on $F$, meaning $\phi(a)=a$, but if $\phi(i)=j \Rightarrow a = \phi(a) = \phi (i^2) = \phi (i)^2 = j^2 = b$, which in not given.
Am I missing something? Where am I thinking wrong?