Let $G =\textrm{GL}_n$, $s \in G$ diagonalizable, $\sigma: G \rightarrow G$ the automorphism $x \mapsto sxs^{-1}$, and $\chi: G \rightarrow G$ the morphism of varieties $x \mapsto sxs^{-1}x^{-1} = (\sigma x)x^{-1}$. Let $G_{\sigma} = \{x \in G: \sigma x = x\}$, and $\mathfrak g_{\sigma} = \textrm{Ker } d \chi$.
It is easy to see that the Lie algebra $L(G_{\sigma})$ of $G_{\sigma}$ is contained in $\mathfrak g_{\sigma}$, and actually, you can show that they are equal. So the dimension of $G_{\sigma}$ is the dimension of $\textrm{Ker } d \chi$.
Given all this information, I'm trying to understand why $T_e \overline{\chi(G)} = \textrm{Im } d \chi$, where $T_e$ denotes the tangent space. This seems to be what is being used in 5.44 in Springer, LAG.
It is clear that '$\supseteq$' holds, and I thought I would do a dimension argument to show equality. We have $$\textrm{Dim Im } d \chi = \textrm{Dim } \mathfrak g - \textrm{Dim Ker } d \chi = \textrm{Dim } G - \textrm{Dim } G_{\sigma}$$
Now I want to say something like $\textrm{Dim } G - \textrm{Dim } G_{\sigma} \geq \textrm{Dim } \chi (G)$, but I'm not sure. Should this be the case?
This is an application of 5.3.2(ii). If $\phi: X \rightarrow Y$ is an equivariant morphism of homogeneous $G$-spaces, $r = \textrm{Dim } Y - \textrm{Dim } X$, $Y'$ is an irreducible component of $Y$, $X'$ is an irreducible component of $\phi^{-1}Y'$, then $\textrm{Dim } X' = \textrm{Dim } Y' +r$.
In particular, if $y \in Y$, then $\textrm{Dim } X' = r$ for any irreducible component $X'$ of $\phi^{-1}\{y\}$.
Let $X = G$ with the group action $g \cdot x = gx$, $Y = \overline{\chi(G)}$ with the action $g \cdot x = (\sigma g)xg^{-1}$, and $\phi = \chi$. The preimage of $\{e\}$ under $\chi$ is $G_{\sigma}$.