Here is a more bare hands explanation. Let $\phi$ be the field automorphism of ${\rm SL}_n(q^2)$ that acts by applying $x \mapsto x^q$ for to the matrix entries. Let $\gamma$ be the graph automorphism that maps matrices $A$ to their inverse-tranpose $A^{- \mathrm{T}}$. Then ${\rm SL}_n(q)$ is the subgroup of ${\rm SL}_n(q^2)$ that is centralized by $\phi$, whereas the group ${\rm SU}_n(q^2)$ (which is confusingly often denoted by ${\rm SU}_n(q)$) that fixes the identity matrix as unitary form is the subgroup of ${\rm SL}_n(q^2)$ that is centralized by $\phi\gamma$.
The automorphism $\gamma$ is outer for $n>2$, but when $n=2$ it is inner and acts in the same way as conjugation by the matrix $\left( \begin{array}{rr}0&1\\ -1&0\end{array} \right)$. It turns out in this case that $\phi$ and $\phi\gamma$ are conjugate in the automorphism group of ${\rm SL}_2(q^2)$ by (the projective image of) an element $g \in {\rm GL}_2(q^2)$, and hence that ${\rm SL}_2(q)$ is conjugate to ${\rm SU}_2(q^2)$ in ${\rm GL}_2(q^2)$. With a bit of calculation on the back of an envelope, we find that $g = \left( \begin{array}{rr}a&b\\ c&d\end{array} \right)$, where $b = -t^qa^q$ and $d= -t^qc^q$ for some field element $t$ with $t^{q+1} = -1$.
Here is a more bare hands explanation. Let $\phi$ be the field automorphism of ${\rm SL}_n(q^2)$ that acts by applying $x \mapsto x^q$ for the matrix entries. Let $\gamma$ be the graph automorphism that maps matrices $A$ to their inverse-tranpose $A^{- \mathrm{T}}$. Then ${\rm SL}_n(q)$ is the subgroup of ${\rm SL}_n(q^2)$ that is centralized by $\phi$, whereas the group ${\rm SU}_n(q^2)$ (which is confusingly often denoted by ${\rm SU}_n(q)$) that fixes the identity matrix as unitary form is the subgroup of ${\rm SL}_n(q^2)$ that is centralized by $\phi\gamma$.
The automorphism $\gamma$ is outer for $n>2$, but when $n=2$ it is inner and acts in the same way as conjugation by the matrix $\left( \begin{array}{rr}0&1\\ -1&0\end{array} \right)$. It turns out in this case that $\phi$ and $\phi\gamma$ are conjugate in the automorphism group of ${\rm SL}_2(q^2)$ by (the projective image of) an element $g \in {\rm GL}_2(q^2)$, and hence that ${\rm SL}_2(q)$ is conjugate to ${\rm SU}_2(q^2)$ in ${\rm GL}_2(q^2)$. With a bit of calculation on the back of an envelope, we find that $g = \left( \begin{array}{rr}a&b\\ c&d\end{array} \right)$, where $b = -t^qa^q$ and $d= -t^qc^q$ for some field element $t$ with $t^{q+1} = -1$.