# An example in which $\dim \operatorname{Aut}(G) < \dim \operatorname{Aut}(\mathfrak g)$

Given a connected Lie group $$G$$ be with Lie algebra $$\mathfrak g$$.

$$\operatorname{Aut}(G)$$ is the group of automorphisms of $$G$$.

$$\operatorname{Aut}(\mathfrak g)$$ is the automorphism group of $$\mathfrak g$$ (which is a Lie group)

It can be proven that the embedding $$\Psi:\operatorname{Aut}(G)\to \operatorname{Aut}(\mathfrak g): f \to T_ef$$ makes $$\operatorname{Aut}(G)$$ into a Lie group.

What would be an example in which $$\dim \operatorname{Aut}(G) < \dim \operatorname{Aut}(\mathfrak g)$$ ?

Let $$G=\mathbb S^1=\{z\in\mathbb C:|z|=1\}$$. Then the only automorphisms of $$G$$ are $$z\mapsto z^{\pm1}$$, so $$\mathrm{Aut}(G)$$ is a finite group.
On the other hand, $$\mathfrak g=i\mathbb R$$ has automorphism group $$\mathrm{Aut}(\mathfrak g)=\mathbb R^\times$$ acting by $$z\mapsto az$$ for $$a\in\mathbb R^\times$$.
Proofs: For the first fact, see How to prove the group of automorphisms of $S^1$ as a topological group is $\mathbb Z_2$?.
For the second fact, note that $$\mathbb R^\times$$ is the automorphism group of $$\mathbb R$$ as a vector space. So the inclusion $$\mathrm{Aut}(\mathfrak g)\subset\mathbb R^\times$$ is clear. Conversely, we need to check that any $$z\mapsto az$$ is a Lie algebra homomorphism. But $$\mathfrak g$$ has trivial Lie bracket so that is clear.