If $L=\mathfrak{sl}(n,F), g\in GL(n,F)$, prove that the map of $L$ to itself defined by $x\rightarrow -gx^tg^{-1}$ ($x^t=$transpose of $x$) belongs to $\operatorname{Aut}L$. When $n=2,g=$identity matrix, prove that this automorphism is inner.
The first part is routine. We check that the map $\varphi$ satisfies $\varphi([x,y])=[\varphi(x),\varphi(y)]:$ $$ -g(xy-yx)^tg^{-1}=gx^ty^tg^{-1}-gy^tx^tg^{-1} $$
so that $\varphi$ is a homomorphism. Moreover, the inverse $\varphi^{-1}$ of $\varphi$ is given by $$\varphi^{-1}(x) = -g^tx(g^{-1})^t$$
so that $\varphi$ is an isomorphism.
Now take $n=2$ and $g=$identity matrix. The map $\varphi$ takes a $2\times 2$ matrix $y$ to $\varphi(y)=-y^t$. We must prove that this automorphism is inner, i.e. it is of the form $\exp\operatorname{ad}x$, with $\operatorname{ad}x$ nilpotent. We have $\exp\operatorname{ad}x=1+\operatorname{ad}x+(\operatorname{ad}x)^2/2!+\ldots$, so that $\exp\operatorname{ad}x(y) = y + [x,y] + [x[x,y]]/2! + \ldots = y + (xy-yx) + \ldots$.
How can we find the right matrix $x$?
Hint: every matrix $y\in\mathfrak{sl}_2(\mathbb K)$ is a unique linear combination $$y=aX+bY+cH,$$ where $(X,Y,H)$ is the standard basis and $a,b,c\in\mathbb K$. If you write $y$ explicitly, you can show that
$$-y^t=aY+bX-cH.$$
The OP asks for a nilpotent matrix $x$ s.t.
$$\exp(\operatorname{ad}_x)(y)=a\exp(\operatorname{ad}_x)(Y)+b\exp(\operatorname{ad}_x)(X)-c\exp(\operatorname{ad}_x)(H) \stackrel{!}{=}aY+bX-cH.$$
This answer here can be useful to finish the proof. Alternatively, what you need is to compute $\exp(\operatorname{ad}_x)(\cdot)$ on the basis of $\mathfrak{sl}_2(\mathbb K)$.