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.
