Let $G = \textrm{GL}_n k$, and let $\sigma: G \rightarrow G$ be an automorphism of algebraic groups. The Lie algebra $\mathfrak g$ of $G$ can be described in three ways:
1 . The space $T_e(G)$ of $k$-derivations of $k[G]$ into $k$, where $k$ is a $k[G]$-module via $f \cdot a = f(e)a$.
2 . The space $L(G)$ consisting of those $k$-derivations $\delta: k[G] \rightarrow k[G]$ which are left invariant: $\lambda_{x^{-1}} \delta \lambda_x = \delta$.
3 . The space of $n$ by $n$ matrices $M_n(k)$.
To go from (1) to (2), define $\psi: T_e(G) \rightarrow L(G)$ by $\psi(\delta)(f)(x) = \delta(\lambda_{x^{-1}}(f))$. To go back from (2) to (1), define $\phi: L(G) \rightarrow T_e(G)$ by $\phi(\delta)(f) = \delta(f)(e)$. To get from (3) to (2), given an $n$ by $n$ matrix $A$, define a derivation $\delta$ on $k[G]$ to $K[G]$ by the formula $\delta(T_{ij}) = - \sum\limits_{h=1}^n T_{ih} a_{hj}$.
Now the corresponding homomorphism of Lie algebras $d \sigma$ can be most easily described if we use the description (1). In that case, $d \sigma(\delta)$ is the derivation $f \mapsto \delta(\sigma^{\ast}(f))$. If we translate $d \sigma$ to a map $L(G) \rightarrow L(G)$, then (according to my computations) that should send a given $\delta \in L(G)$ to the derivation which sends any $f \in k[G]$ to the regular function given by $$ x \mapsto \delta(\sigma^{\ast}(\lambda_{x^{-1}}(f)))(e) $$ If $\sigma^{\ast}$ commutes with left translation then the formula is simpler: $d \sigma: L(G) \rightarrow L(G)$ is given by $d \sigma(\delta)(f) = \delta(\sigma^{\ast}(f))$.
If we identify the Lie algebra of $G$ with $M_n(k)$, is there a nice way to get a formula for $d \sigma: M_n(k) \rightarrow M_n(k)$ in general? Or does one need to go through the identifications in (1), (2), (3) to derive an explicit formula every time?
For example, I tried letting $\sigma$ be the map $x \mapsto sxs^{-1}$, where $$s = \begin{pmatrix} \lambda_1 & & & 0 \\ & \lambda_2 \\ & & \ddots \\ 0 & & & \lambda_n \end{pmatrix}$$ is a fixed invertible diagional matrix. I worked through the formulas above and got $d \sigma: M_n(k) \rightarrow M_n(k)$ as the map which sends any matrix $(a_{ij})$ to $(\lambda_i \lambda_j^{-1}a_{ij})$.
Basically, you get something like $\sigma(I_n + \varepsilon \cdot M) = I_n + \varepsilon \cdot d\sigma(M)$ mod $\varepsilon^2$ which is the usual intuition of what a differential is. To make rigourous sense of this you can say that you take $k[\varepsilon]$-valued points of the group, with $\varepsilon^2 = 0$.
So if $\sigma$ is inner, then it extends to an linear automorphism of $M_n(k)$ (and even $M_n(k[\varepsilon])$) and $\sigma(I_n + \varepsilon \cdot M) = I_n +\varepsilon\cdot \sigma(M)$ so $d\sigma(M) = \sigma(M)$.