Question about calculation of a vector field on a Lie group Let $G$ be a compact Lie group with Lie algebra $\mathfrak{g}$.
Let $g \in G$ and $X \in \mathfrak{g}$ such that
$$\frac{d}{dt} \vert_{t=0} ge^{tX}g^{-1} = X.$$
What can we say about
$$\frac{d}{dt} \vert_{t=0} e^{-tX} g e^{tX}  \qquad ?$$
 A: *

*Getting my head around your first equation
The easiest is to consider $G=SO(3)$ and $\mathfrak{g}$ the space of traceless antisymmetric $3\times 3$-matrices
$$
X=\left(\begin{matrix}0&-x_3&x_2\\x_3&0&-x_1\\-x_2&x_1&0\end{matrix}\right)
$$
(which is isomorphic to $\mathbb R^3)\,.$
In general,
$$
\frac{d}{dt}\Bigg|_{t=0}\,g\,e^{tX}\,g^{-1}=gXg^{-1}\,.
$$
So, your first equation (with sign corrected)
$$\tag{1}
\frac{d}{dt}\Bigg|_{t=0}\,g\,e^{tX}\,g^{-1}=X
$$
means nothing else than $gXg^{-1}=X\,,$ or,
$$
gX=Xg\,.
$$
In other words, the rotation matrix $g$ and the matrix $X$ commute.
The rest of this section can be skipped.
For $\boldsymbol{v}\in\mathbb R^3$ we also have obviously
$$
X\boldsymbol{v}=\boldsymbol{x}\times \boldsymbol{v}
$$
where $\boldsymbol{x}=(x_1,x_2,x_3)^\top\,,\boldsymbol{v}=(v_1,v_2,v_3)^\top\,.$ Therefore, $gX\boldsymbol{v}=Xg\,\boldsymbol{v}$ can be written as
$$
g(\boldsymbol{x}\times \boldsymbol{v})=\boldsymbol{x}\times (g\,\boldsymbol{v})\,.
$$
Because this must hold for all $\boldsymbol{v}$ it must hold for the rotation axis of $g$ so that in this case
$$\tag{2}
g(\boldsymbol{x}\times \boldsymbol{v})=\boldsymbol{x}\times \boldsymbol{v}\,.
$$
This means that only the following case is possible:

*

*$\boldsymbol{x}$ is parallel to the rotation axis of $g$ (in which case $\boldsymbol{x}\times\boldsymbol{v}=\boldsymbol{0}$)

Note that if $\boldsymbol{x}\times\boldsymbol{v}\not=\boldsymbol{0}$ it must be parallel to the rotation axis of $g$ (by (2)). But it is not possible that
$\boldsymbol{x}\times\boldsymbol{v}\not=\boldsymbol{0}$ and $\boldsymbol{v}\not=\boldsymbol{0}$
are parallel.


*Your second equation
By the product rule,
\begin{align}
\frac{d}{dt}\Bigg|_{t=0}\,\Big(e^{-tX}\,g\,e^{tX}\Big)&=\frac{d}{dt}\Bigg|_{t=0}\,\Big(e^{-tX}\,g\Big)\,e^{tX}\Bigg|_{t=0}+e^{-tX}\Bigg|_{t=0}g\frac{d}{dt}\Bigg|_{t=0}\,e^{tX}\\[3mm]
&=-Xg+gX\tag{3}
\end{align}
which is just the commutator $[g,X]\,.$
Therefore, if you assume that $g$ and $X$ satisfy the first equation the second equation will be zero.
A: If the first equation holds, the second expression will vanish.
There's actually no loss of generality in assuming that $G$ is a Matrix Lie group, since the computation is local, and every Lie group is locally equivalent to a matrix Lie group. As Kurt G. point out, in this case the first equation simplifies to $gXg^{-1}=X$ and the second expression simplifies to $-Xg+gX$.
If you want to do things in terms of abstract Lie groups, that is also possible, but some more machinery is needed. We can use the standard identification $\mathfrak{g}\cong T_eG$, the left and right translation operators $L_g(h):=gh$ and $R_g(h):=hg$, the adjoint representation $\operatorname{Ad}_g=d_e(L_g\circ R_{g^{-1}}):\mathfrak{g}\to\mathfrak{g}$, and the following product rule:
$$
\frac{d}{dt}(a(t)b(t))=dR_{b(t)}\left(a'(t)\right)+dL_{a(t)}\left(b'(t)\right)
$$
Where $a,b:\mathbb{R}\to G$ are paths in $G$. Putting these all together, we have
$$
\frac{d}{dt}\left(ge^{tX}g^{-1}\right)_{t=0}=dL_g(dR_{g^{-1}}(X))=\operatorname{Ad}_g(X)
$$
$$
\frac{d}{dt}\left(e^{-tX}ge^{tX}\right)_{t=0}=dR_g(-X)+dL_g(X)=dL_g(X-\operatorname{Ad}_g(X))
$$
