[proofplan]
We use that the adjoint action of a Lie group element is a [Lie algebra](/page/Lie%20Algebra) automorphism. This gives an intertwining identity between the infinitesimal adjoint maps $\operatorname{ad}_X$ and $\operatorname{ad}_{\operatorname{Ad}_gX}$. Substituting this identity into the trace formula defining the Killing form turns $B(\operatorname{Ad}_gX,\operatorname{Ad}_gY)$ into the trace of a conjugate of $\operatorname{ad}_X\circ\operatorname{ad}_Y$, and conjugation invariance of trace gives the desired equality.
[/proofplan]
[step:Intertwine the infinitesimal adjoint maps by the adjoint representation]
Fix $g\in G$. Define the conjugation map
\begin{align*}
C_g:G\to G,\qquad h\mapsto ghg^{-1}.
\end{align*}
By the definition of the adjoint representation [citetheorem:8821], $\operatorname{Ad}_g=d(C_g)_e:\mathfrak g\to\mathfrak g$. Since $C_g$ is a Lie group automorphism with inverse $C_{g^{-1}}$, its differential $\operatorname{Ad}_g$ is a linear isomorphism with inverse $\operatorname{Ad}_{g^{-1}}$.
The [differential of a Lie group homomorphism preserves Lie brackets](/theorems/8803) [citetheorem:8803]. Applying this to the automorphism $C_g:G\to G$, we obtain, for all $U,V\in\mathfrak g$,
\begin{align*}
\operatorname{Ad}_g[U,V]=[\operatorname{Ad}_gU,\operatorname{Ad}_gV].
\end{align*}
Let $X\in\mathfrak g$. We claim that
\begin{align*}
\operatorname{ad}_{\operatorname{Ad}_gX}=\operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{Ad}_{g^{-1}}.
\end{align*}
Indeed, for every $Z\in\mathfrak g$,
\begin{align*}
(\operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{Ad}_{g^{-1}})(Z)=\operatorname{Ad}_g[X,\operatorname{Ad}_{g^{-1}}Z].
\end{align*}
Using bracket preservation with $U=X$ and $V=\operatorname{Ad}_{g^{-1}}Z$ gives
\begin{align*}
\operatorname{Ad}_g[X,\operatorname{Ad}_{g^{-1}}Z]=[\operatorname{Ad}_gX,\operatorname{Ad}_g\operatorname{Ad}_{g^{-1}}Z].
\end{align*}
Since $\operatorname{Ad}_{g^{-1}}$ is the inverse of $\operatorname{Ad}_g$, this equals
\begin{align*}
[\operatorname{Ad}_gX,Z]=\operatorname{ad}_{\operatorname{Ad}_gX}(Z).
\end{align*}
Thus the two linear maps agree on every $Z\in\mathfrak g$.
[guided]
Fix $g\in G$. The point of this step is to translate the group-level conjugation action into a statement about the linear maps whose traces define the Killing form. Define
\begin{align*}
C_g:G\to G,\qquad h\mapsto ghg^{-1}.
\end{align*}
This is a smooth Lie group automorphism, with inverse $C_{g^{-1}}:G\to G$. By the definition of the adjoint representation [citetheorem:8821], its differential at the identity is
\begin{align*}
\operatorname{Ad}_g=d(C_g)_e:\mathfrak g\to\mathfrak g.
\end{align*}
Since $C_{g^{-1}}$ is the inverse of $C_g$, the chain rule gives that $\operatorname{Ad}_{g^{-1}}$ is the inverse [linear map](/page/Linear%20Map) of $\operatorname{Ad}_g$.
We now use the fact that differentials of Lie group homomorphisms preserve Lie brackets [citetheorem:8803]. The map $C_g:G\to G$ is a Lie [group homomorphism](/page/Group%20Homomorphism) because, for all $a,b\in G$,
\begin{align*}
C_g(ab)=gabg^{-1}=(gag^{-1})(gbg^{-1})=C_g(a)C_g(b).
\end{align*}
Therefore its differential preserves brackets: for all $U,V\in\mathfrak g$,
\begin{align*}
\operatorname{Ad}_g[U,V]=[\operatorname{Ad}_gU,\operatorname{Ad}_gV].
\end{align*}
We want an identity involving $\operatorname{ad}_{\operatorname{Ad}_gX}$, which is the map that brackets with $\operatorname{Ad}_gX$. Let $X\in\mathfrak g$ and $Z\in\mathfrak g$. Because $\operatorname{Ad}_{g^{-1}}$ is the inverse of $\operatorname{Ad}_g$, we may first pull $Z$ back to $\operatorname{Ad}_{g^{-1}}Z$, bracket with $X$, and then push forward by $\operatorname{Ad}_g$:
\begin{align*}
(\operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{Ad}_{g^{-1}})(Z)=\operatorname{Ad}_g[X,\operatorname{Ad}_{g^{-1}}Z].
\end{align*}
Applying bracket preservation with $U=X$ and $V=\operatorname{Ad}_{g^{-1}}Z$ gives
\begin{align*}
\operatorname{Ad}_g[X,\operatorname{Ad}_{g^{-1}}Z]=[\operatorname{Ad}_gX,\operatorname{Ad}_g\operatorname{Ad}_{g^{-1}}Z].
\end{align*}
The last factor simplifies because $\operatorname{Ad}_g\operatorname{Ad}_{g^{-1}}=\operatorname{id}_{\mathfrak g}$, so
\begin{align*}
[\operatorname{Ad}_gX,\operatorname{Ad}_g\operatorname{Ad}_{g^{-1}}Z]=[\operatorname{Ad}_gX,Z].
\end{align*}
By the definition of $\operatorname{ad}_{\operatorname{Ad}_gX}$, this final expression is
\begin{align*}
[\operatorname{Ad}_gX,Z]=\operatorname{ad}_{\operatorname{Ad}_gX}(Z).
\end{align*}
Since the equality holds for every $Z\in\mathfrak g$, we have proved the linear-map identity
\begin{align*}
\operatorname{ad}_{\operatorname{Ad}_gX}=\operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{Ad}_{g^{-1}}.
\end{align*}
[/guided]
[/step]
[step:Substitute the intertwining identities into the trace defining the Killing form]
Let $X,Y\in\mathfrak g$. By the definition of the Killing form,
\begin{align*}
B(\operatorname{Ad}_gX,\operatorname{Ad}_gY)=\operatorname{tr}(\operatorname{ad}_{\operatorname{Ad}_gX}\circ\operatorname{ad}_{\operatorname{Ad}_gY}).
\end{align*}
Applying the intertwining identity from the previous step to both $X$ and $Y$ gives
\begin{align*}
\operatorname{ad}_{\operatorname{Ad}_gX}\circ\operatorname{ad}_{\operatorname{Ad}_gY}=(\operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{Ad}_{g^{-1}})\circ(\operatorname{Ad}_g\circ\operatorname{ad}_Y\circ\operatorname{Ad}_{g^{-1}}).
\end{align*}
Using associativity of composition and $\operatorname{Ad}_{g^{-1}}\circ\operatorname{Ad}_g=\operatorname{id}_{\mathfrak g}$, this composition is
\begin{align*}
\operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{ad}_Y\circ\operatorname{Ad}_{g^{-1}}.
\end{align*}
Hence
\begin{align*}
B(\operatorname{Ad}_gX,\operatorname{Ad}_gY)=\operatorname{tr}(\operatorname{Ad}_g\circ\operatorname{ad}_X\circ\operatorname{ad}_Y\circ\operatorname{Ad}_{g^{-1}}).
\end{align*}
[/step]
[step:Use conjugation invariance of trace to recover the original Killing form]
Set
\begin{align*}
T:=\operatorname{ad}_X\circ\operatorname{ad}_Y:\mathfrak g\to\mathfrak g.
\end{align*}
Since $\mathfrak g$ is finite-dimensional and $\operatorname{Ad}_g:\mathfrak g\to\mathfrak g$ is an invertible linear map with inverse $\operatorname{Ad}_{g^{-1}}$, the elementary trace identity
\begin{align*}
\operatorname{tr}(STS^{-1})=\operatorname{tr}(T)
\end{align*}
applies with $S=\operatorname{Ad}_g$; this follows from cyclicity of trace, because $\operatorname{tr}(STS^{-1})=\operatorname{tr}(TS^{-1}S)=\operatorname{tr}(T)$. Therefore
\begin{align*}
\operatorname{tr}(\operatorname{Ad}_g\circ T\circ\operatorname{Ad}_{g^{-1}})=\operatorname{tr}(T).
\end{align*}
Substituting the definition of $T$ gives
\begin{align*}
B(\operatorname{Ad}_gX,\operatorname{Ad}_gY)=\operatorname{tr}(\operatorname{ad}_X\circ\operatorname{ad}_Y)=B(X,Y).
\end{align*}
Since $g\in G$ and $X,Y\in\mathfrak g$ were arbitrary, the Killing form is invariant under the adjoint action of $G$.
[/step]
