Let $\mathfrak{G}$ be a Lie group with algebra $\mathfrak{g}$ and with in general many connected components but which is either centerless or has at most a discrete center.
Then we have an "indempotence" for $\mathrm{ad}:\mathfrak{g}\to\mathfrak{gl}(\mathfrak{g})$ insofar that we have $\mathrm{ad}(\mathrm{ad}(X)) = \mathrm{ad}(X)\,\forall\,X\in\mathfrak{g}$. $^1$
Moreover, the identity connected components of $\mathrm{Ad}(\mathrm{Ad}(\mathfrak{G}))$ and of $\mathrm{Ad}(\mathfrak{G})$ are the same (think of them as being generated by exponentials of the same matrices).
However, it is conceivable that $\mathrm{Ad}$ can still map a noncentral $\gamma\in\mathfrak{G}$ that defines a nonidentity connected component $\gamma\,\mathfrak{G}$ so that $\mathrm{Ad}(\gamma)$ becomes central in $\mathrm{Ad}(\mathfrak{G})$. This would happen if the commutators of noncentral $\gamma$ and all other $\mathfrak{G}$ members were central. In this case, the second application of $\mathrm{Ad}$ would merge connected component $\mathrm{Ad}(\gamma)\,\mathrm{Ad}(\mathfrak{G})$ into the identity component, so the second application of $\mathrm{Ad}$ may not be an isomorphism of the whole group $\mathrm{Ad}(\mathfrak{G})$.
Is this possible? If so, can you give an example? If not, why?
$^1$: $\mathrm{ad}$ having trivial kernel is an isomorphism of Lie algebras, so if $X\in\mathfrak{g}$, the matrix of $\mathrm{ad}(X)$ encodes the Lie brackets between $X$ and other members of $\mathfrak{g}$, and, by dint of the Jacobi identity $\mathrm{ad}([X,\,Y])=[\mathrm{ad}(X),\,\mathrm{ad}(Y)]$ $^2$ these are the same bracket relationships as those between $\mathrm{ad}(X)$ and the corresponding members of $\mathfrak{g}$. So we have $\mathrm{ad}(\mathrm{ad}(X)) = \mathrm{ad}(X)\,\forall\,X\in\mathfrak{g}$.
$^2$:yes, although it's the statement that $\mathrm{ad}$ is a Lie algebra homomorphism, but it's the Jacobi identity in disguise