If $=(,)$,$∈(,)$, prove that the map $L$ to itself defined by $\sigma(z)=-gz^tg$ belongs to Aut$(L)$. When $n=2$, $g$=identity matrix, prove that this automorphism is inner.
The first part is clear to me. When $n=2$,and $g$ is the identity matrix, it is easy to see that $\sigma(x)=-y,\sigma(y)=-x, \sigma(h)=-h$, where $x=\begin{bmatrix}0&1\\0&0\end{bmatrix}$,$y=\begin{bmatrix}0&0\\1&0\end{bmatrix}$, $h=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ comprise the basis of $L$. I was looking at the answers here (Automorphism in the special linear algebra $\mathfrak{sl}_2(F)$, Inner automorphism of $\mathfrak{sl}(2,k)$, $\operatorname{char}(k)=0$ and adjoint action) but I still don't understand a couple of things. How is $(adX)^n=0$ for $n\geq 3$ when $X$ is nilpotent? Also, how can we write $\sigma$=exp(ad$x$)expad$(-y)$expad$x$, thereby concluding that $\sigma$ is an inner automorphism?
