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Complementing Jose's answer: let $G$ be a complex semi-simple simply connected Lie group and let $g$ be the Lie algebra of $G$. The outer automorphism of $g$ is the automorphism group of the Dynkin diagram. Briefly, given an automorphism, we can assume that it preserves a given Cartan subalgebra (or else multiply by an element of the form $Ad(g), g\in G$ that takes one Cartan subalgebra to another one; since the exponential is surjective [edit: no it isn't, as pointed out by Theo, but it is locally] and $Ad(exp(x))=exp(ad(x)),x\in g$, this is an inner automorphism [edit: this only works for $g$ sufficiently close to the unit; in general write $g$ as a product of the exponentials and apply this to each factor]). Any automorphism preserving a Cartan subalgebra $h$ induces an automorphism of the root system, and all automorphisms of the root system arise in this way. Moreover, the automorphisms that induce the identical mapping of the root system are precisely those of the form $exp(ad(x)),x\in h$ (this requires a little check but is not massively difficult).

Now, since complexifying and taking the Lie algebra induces an equivalence of categories of compact simply connected semi-simple Lie groups and complex semi-simple Lie algebras, the above conclusion holds for the automorphism groups of compact simply connected semi-simple Lie groups as well: namely, the outer automorphism group is the automorphism group of the root system (or, which is the same, the automorphism group of the Dynkin diagram).

Complementing Jose's answer: let $G$ be a complex semi-simple simply connected Lie group and let $g$ be the Lie algebra of $G$. The outer automorphism of $g$ is the automorphism group of the Dynkin diagram. Briefly, given an automorphism, we can assume that it preserves a given Cartan subalgebra (or else multiply by an element of the form $Ad(y), g\in G$ that takes one Cartan subalgebra to another one; since the exponential is surjective [edit: no it isn't, as pointed out by Theo, but it is locally] and $Ad(exp(x))=exp(ad(x)),x\in g$, this is an inner automorphism [edit: this only works for $g$ sufficiently close to the unit; in general write $y$ as a product of the exponentials and apply this to each factor]). Any automorphism preserving a Cartan subalgebra $h$ induces an automorphism of the root system, and all automorphisms of the root system arise in this way. Moreover, the automorphisms that induce the identical mapping of the root system are precisely those of the form $exp(ad(x)),x\in h$ (this requires a little check but is not massively difficult).

Now, since complexifying and taking the Lie algebra induces an equivalence of categories of compact simply connected semi-simple Lie groups and complex semi-simple Lie algebras, the above conclusion holds for the automorphism groups of compact simply connected semi-simple Lie groups as well: namely, the outer automorphism group is the automorphism group of the root system (or, which is the same, the automorphism group of the Dynkin diagram).

Complementing Jose's answer: let $G$ be a complex semi-simple simply connected Lie group and let $g$ be the Lie algebra of $G$. The outer automorphism of $g$ is the automorphism group of the Dynkin diagram. Briefly, given an automorphism, we can assume that it preserves a given Cartan subalgebra (or else multiply by an element of the form $Ad(g), g\in G$ that takes one Cartan subalgebra to another one; since the exponential is surjective and $Ad(exp(x))=exp(ad(x)),x\in g$, this is an inner automorphism). Any automorphism preserving a Cartan subalgebra $h$ induces an automorphism of the root system, and all automorphisms of the root system arise in this way. Moreover, the automorphisms that induce the identical mapping of the root system are precisely those of the form $exp(ad(x)),x\in h$ (this requires a little check but is not massively difficult).

Now, since complexifying and taking the Lie algebra induces an equivalence of categories of compact simply connected semi-simple Lie groups and complex semi-simple Lie algebras, the above conclusion holds for the automorphism groups of compact simply connected semi-simple Lie groups as well: namely, the outer automorphism group is the automorphism group of the root system (or, which is the same, the automorphism group of the Dynkin diagram).

Complementing Jose's answer: let $G$ be a complex semi-simple simply connected Lie group and let $g$ be the Lie algebra of $G$. The outer automorphism of $g$ is the automorphism group of the Dynkin diagram. Briefly, given an automorphism, we can assume that it preserves a given Cartan subalgebra (or else multiply by an element of the form $Ad(g), g\in G$ that takes one Cartan subalgebra to another one; since the exponential is surjective [edit: no it isn't, as pointed out by Theo, but it is locally] and $Ad(exp(x))=exp(ad(x)),x\in g$, this is an inner automorphism [edit: this only works for $g$ sufficiently close to the unit; in general write $g$ as a product of the exponentials and apply this to each factor]). Any automorphism preserving a Cartan subalgebra $h$ induces an automorphism of the root system, and all automorphisms of the root system arise in this way. Moreover, the automorphisms that induce the identical mapping of the root system are precisely those of the form $exp(ad(x)),x\in h$ (this requires a little check but is not massively difficult).

Now, since complexifying and taking the Lie algebra induces an equivalence of categories of compact simply connected semi-simple Lie groups and complex semi-simple Lie algebras, the above conclusion holds for the automorphism groups of compact simply connected semi-simple Lie groups as well: namely, the outer automorphism group is the automorphism group of the root system (or, which is the same, the automorphism group of the Dynkin diagram).

Complementing Jose's answer: let $G$ be a complex semi-simple simply connected Lie group and let $g$ be the Lie algebra of $G$. The outer automorphism of $g$ is the automorphism group of the Dynkin diagram. Briefly, given an automorphism, we can assume that it preserves a given Cartan subalgebra (or else multiply by an element of the form $Ad(g)$ that takes one Cartan subalgebra to another one; since the exponential is surjective and $Ad(exp(x))=exp(ad(x))$, this is an inner automorphism). Any automorphism preserving a Cartan subalgebra $h$ induces an automorphism of the root system, and all automorphisms of the root system arise in this way. Moreover, the automorphisms that induce the identical mapping of the root system are precisely those of the form $exp(ad(x)),x\in H$ (this requires a little check but is not massively difficult).

Now, since complexifying and taking the Lie algebra induces an equivalence of categories of compact simply connected semi-simple Lie groups and complex semi-simple Lie algebras, the above conclusion holds for the automorphism groups of compact simply connected semi-simple Lie groups as well: namely, the outer automorphism group is the automorphism group of the root system (or, which is the same, the automorphism group of the Dynkin diagram).

Complementing Jose's answer: let $G$ be a complex semi-simple simply connected Lie group and let $g$ be the Lie algebra of $G$. The outer automorphism of $g$ is the automorphism group of the Dynkin diagram. Briefly, given an automorphism, we can assume that it preserves a given Cartan subalgebra (or else multiply by an element of the form $Ad(g), g\in G$ that takes one Cartan subalgebra to another one; since the exponential is surjective and $Ad(exp(x))=exp(ad(x)),x\in g$, this is an inner automorphism). Any automorphism preserving a Cartan subalgebra $h$ induces an automorphism of the root system, and all automorphisms of the root system arise in this way. Moreover, the automorphisms that induce the identical mapping of the root system are precisely those of the form $exp(ad(x)),x\in h$ (this requires a little check but is not massively difficult).

Now, since complexifying and taking the Lie algebra induces an equivalence of categories of compact simply connected semi-simple Lie groups and complex semi-simple Lie algebras, the above conclusion holds for the automorphism groups of compact simply connected semi-simple Lie groups as well: namely, the outer automorphism group is the automorphism group of the root system (or, which is the same, the automorphism group of the Dynkin diagram).