Denote by $G$ a Lie group with corresponding Lie algebra $\text{Lie}(G)$.
There the three maps
inner automorphism/conjugation: $\text{Int}_g = L_{g^{-1}} \circ R_g \in \text{Aut}(G)$,
$\text{Ad}_g := D_1 \text{Int}_g = D_g L_{g^{-1}} \circ D_1 R_g \in \text{Aut}(\text{Lie}(G))$,
and moreover: $\text{ad}_v := \frac{d}{dt}\big\vert_{t=0} \text{Ad}_{\gamma_v(t)}\in\text{Hom}(\text{Lie}(G))$ where $\gamma_v(t)$ is any one-parameter semi-group generated by $v$.
Now, in my lecture there it says ($G$ is not said to be embedded in a matrix):
The component $K$ of identity of $G$ is normal, i.e. it is invariant under $\text{Ad}_g$ for any $g\in G$.
Since normality is defined by $k=g^{-1} k g$ for any $g\in G$ and all $k\in K$ this statement would imply that $\text{Int}_g = \text{Ad}_g$ on $K$.
Is this true? I do not see how to see that $\text{Ad}_g$ behaves there like this.