When does the automorphism group of a Lie Group equal to the automorphism group of its Lie Algebra

Given that $$G$$ is a connected Lie Group and $$\mathfrak{g}$$ is its Lie Algebra. Denote $$\tilde{G}$$ as the universal covering Lie group of $$G$$. We know that $$\mathrm{Aut}\left(\tilde{G}\right)=\mathrm{Aut}\left(\mathfrak{g}\right)$$. We also know that $$\mathrm{Aut}\left(G\right)$$ is a closed subgroup of $$\mathrm{Aut}\left(\tilde{G}\right)$$, hence it's a Lie subgroup of $$\mathrm{Aut}\left(\mathfrak{g}\right)$$. Also We know that $$\mathrm{Aut}\left(G\right)=\mathrm{Aut}\left(\mathfrak{g}\right)$$ if $$G$$ is simply connected, as in such case we have $$G=\tilde{G}$$.
My question is, does the conclusion works from the other side? i.e.
For a connected (compact if needed) Lie group $$G$$, if $$\mathrm{Aut}\left(G\right)=\mathrm{Aut}\left(\mathfrak{g}\right)$$, do we know for sure that $$G$$ is simply connected?

• What non-simply connected $G$ have you checked as examples? Commented May 27, 2022 at 15:05
• $G=S^{2}$ for example, in which $\mathrm{Aut}\left(G\right)=\mathbb{Z}_{2}$ while $\mathrm{Aut}\left(\mathfrak{g}\right)=\mathbb{R}^{*}$. Commented May 27, 2022 at 15:36
• Sorry, It's a typo here. It should be $S^{1}$ instead of $S^{2}$. Yet the rest is correct. Commented May 27, 2022 at 16:19
• I thought that's a typo ... Good. What about some simple group like $SL_2(\mathbb R)$ or $SL_2(\mathbb C)$, or, for the compact case, $SU_2$? Commented May 27, 2022 at 16:48
• You might want to check out the excellent answer to math.stackexchange.com/q/4419869/96384 (which contains far more info, but you should distill what you need in case $K= \mathbb R, \mathbb C$). As for $SU_2$, that is actually simply connected, so is irrelevenat for your question; instead you want to look at $SO_n$ and the answers (plus discussion in comments) here: math.stackexchange.com/q/3104437/96384 Commented May 27, 2022 at 17:18