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Sorry, I'm not a specialist, I want to ask about automorphisms of the group $SO(n,\mathbb{R})$ ($\mathbb{R}$ - field of reals). It is easy that a function of the form $f_C(A)=CAC^{-1}$ for $A \in SO(n, \mathbb{R})$, where $C\in O(n,\mathbb{R})$, is an automorphism.

But, is it true that each automorphism of $SO(n,\mathbb{R})$ is of the form $f_C$ with $C \in O(n,\mathbb{R})$ or maybe with $C \in SO(n,\mathbb{R})$?

Thanks.

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    Sorry, I have just corrected.2011-08-26

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See Outer automorphism group wiki page, the section on real Lie groups. It says that outer automorphism groups are symmetries of Dynkin diagram.

From this it follows that for $SO(2n-1, \mathbb{R})$, i.e. series $B_n$, all automorphisms are inner. For $SO(2n, \mathbb{R})$ there is order 2 outer automorphism which indeed coincides with conjugation by reflections.

So it follows that the answer to your question is in affirmative, and $C \in SO(n, \mathbb{R})$ for odd $n$, and in $O(n, \mathbb{R})$ for even.

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    To continue what @Sashha wrote,$SO(8)$does not admit an automorphism corresponding to the "triality" automorphism of the D_4 Dynkin diagram. The group$Spin(8)$does, but it is quite magical. In particular, the usual construction of$Spin(8)$embeds it as a subgroup of the Clifford algebra Cliff(8). The triality automorphism *does not* extend to a linear automorphism of Cliff(8). Googling for "triality" will find descriptions.2016-03-15