The experts should correct me if there is a fatal mistake in the argument, I am neither an algebraic geometer nor a Lie theorist. I am working over $\mathbb{C}$.
1. Let $\mathfrak{g}$ be a semisimple Lie algebra, $G$ be the adjoint group, $Aut(\mathfrak{g})$ be the automorphism group. Let $Aut_0 (\mathfrak{g})$ be the subgroup of automorphisms that preserve the decomposition of $\mathfrak{g}$ into simple ideals. It has finite index in the whole automorphism group, since the decomposition into ideals is unique and an automorphism has to map a simple ideal into a simple ideal. The group $Aut_0(\mathfrak{g}$ is the product of the automorphism groups of the simple factors. The automorphism group of any simple Lie algebra has the adjoint group as a finite index subgroup; the quotient is the automorphism group of the Dynkin diagram. The upshot of this discussion is: $Aut(\mathfrak{g})$ has $G$ as a finite index subgroup. In particular, the dimension of $Aut(\mathfrak{g})$ only depends on the dimension of $\mathfrak{g}$!
1. Let $V$ be the variety of all Lie algebra structures on $C^n$; it has an action of $GL_n$ on it, the stabilizers are the automorphism groups, the orbits are the sets of isomorphic Lie algebrasisomorphism classes. Now let $\mathfrak{g} \in V$ be semisimple and let $O \subset V$ be its $GL_n$-orbit; this is the set of all Lie algebras isomorphic to $\mathfrak{g}$. GL_n$-orbit. Now use the orbit closure theorem (Borel, Linear Algebraic Groups, page 53). It says that any$GL_n$-orbit$O$is open in its closure and that$\bar{O} \setminus O$consists of orbits of smaller dimension. Hence all Lie algebras in the closure of$O$which are not isomorphic to$\mathfrak{g}$must have a larger-dimensional automorphism group. By part 1, they are not semisimple. 1 The experts should correct me if there is a fatal mistake in the argument, I am neither an algebraic geometer nor a Lie theorist. I am working over$\mathbb{C}$. 1. Let$\mathfrak{g}$be a semisimple Lie algebra,$G$be the adjoint group,$Aut(\mathfrak{g})$be the automorphism group. Let$Aut_0 (\mathfrak{g})$be the subgroup of automorphisms that preserve the decomposition of$\mathfrak{g}$into simple ideals. It has finite index in the whole automorphism group, since the decomposition into ideals is unique and an automorphism has to map a simple ideal into a simple ideal. The group$Aut_0(\mathfrak{g}$is the product of the automorphism groups of the simple factors. The automorphism group of any simple Lie algebra has the adjoint group as a finite index subgroup; the quotient is the automorphism group of the Dynkin diagram. The upshot of this discussion is:$Aut(\mathfrak{g})$has$G$as a finite index subgroup. In particular, the dimension of$Aut(\mathfrak{g})$only depends on the dimension of$\mathfrak{g}$! 2. Let$V$be the variety of all Lie algebra structures on$C^n$; it has an action of$GL_n$on it, the stabilizers are the automorphism groups, the orbits are the sets of isomorphic Lie algebras. Now let$\mathfrak{g} \in V$be semisimple and let$O \subset V$be its$GL_n$-orbit; this is the set of all Lie algebras isomorphic to$\mathfrak{g}$. Now use the orbit closure theorem (Borel, Linear Algebraic Groups, page 53). It says that any$GL_n$-orbit$O$is open in its closure and that$\bar{O} \setminus O$consists of orbits of smaller dimension. Hence all Lie algebras in the closure of$O$which are not isomorphic to$\mathfrak{g}\$ must have a larger-dimensional automorphism group. By part 1, they are not semisimple.