>If $\mathfrak{s}$ is a simple Lie algebra over an algebraically closed field of characteristic zero or a classical simple Lie algebra over an algebraically closed field of characteristic $p>3$, then the trace over the automorphism group $\operatorname{Aut}(\mathfrak{s})$ is surjective.

Let $\mathfrak{h}$ be a Cartan subalgebra and $\Phi$ its root system, we have the decomposition 
$$\mathfrak{s}=\mathfrak{h}\oplus\bigoplus_{\alpha\in\Phi}\mathfrak{s}_\alpha$$
where $\mathfrak{s}_\alpha=\{x\in\mathfrak{s}\vert~ [h,x]=\alpha(h)x\text{ for all }h\in\mathfrak{h}\}$ is of dimension $1$. For a positive root $\gamma$, we set $\Lambda_\gamma=\{\beta\in\Phi\vert~\gamma+\beta\in\Phi\}$ and $\Lambda_\gamma^\pm=\Lambda_\gamma\cap\Phi^\pm$. Let $c_\gamma$ be a nonzero scalar, with respect to the Chevalley basis $\{x_\alpha,h_i\vert~\alpha\in\Phi\text{ and  }1\leq i\leq\dim\mathfrak{h}\}$, we define the automorphism $\varphi_\gamma$ as follows,
\begin{align*}
    \left\{\begin{array}{ll}
        \varphi_\gamma(h)=h,&\text{for all } h\in\mathfrak{h},\\
         \varphi_\gamma(x_\beta)=c_\gamma^{\pm 1} x_\beta,&\text{if } \beta\in\Lambda_\gamma^\pm, \\
         %\phi(x_\alpha)=c_\alpha^{-1} x_\alpha,&~\text{if } \\
         \varphi_\gamma(x_\beta)= x_\beta,&\text{otherwise}.
    \end{array}
    \right.
\end{align*}
Hence 
\begin{align*}
    \operatorname{tr}(\varphi_\gamma)&=\dim\mathfrak{h}+\vert\Phi\vert-\vert\Lambda_\gamma\vert+c_\gamma\vert\Lambda^+_\alpha\vert+c_\gamma^{-1}\vert\Lambda_\gamma^{-}\vert\\
    &=\dim\mathfrak{s}+\left(\frac{c_\gamma+c_\gamma^{-1}}{2}-1\right)\vert\Lambda_\gamma\vert
\end{align*}