## Return to Answer

2 deleted 1 characters in body

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 $t^{q+1} = -1$.

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 $t^{q+1} = -1$.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 54, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9684198498725891, "perplexity_flag": "head"}