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I would like to know what the group of outer automorphisms of $Sp(2)$ is. I think this should be isomorphic to $\mathbb{Z}_2$, but I am not completely sure.

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    No the corresponding Dynkin diagram (of type $B_2=C_2$) has no symmetry, Out is trivial. – YCor Jun 09 '15 at 22:50
  • But, one could consider for instance, the map sending a $2 \times 2$ matrix $a=(a_{ij})$ in $\Sp(2)$ to $b=(b_{ij})$ with $a_{ii}=b_{ii}$ and $b_{ij}=a_{ji}$. Isn't this an outer automorphism? –  Jun 09 '15 at 22:55
  • I'm not sure what you call $Sp(2)$ then... – YCor Jun 09 '15 at 23:08
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    I am sorry if the notation is confusing. Other notation for what I mean is $USp(4)=U(4) \cap Sp(4, \mathbb{C})$. You probably thought I meant $Sp(2,\mathbb{C})$, case in which your observation is clear. –  Jun 09 '15 at 23:14
  • As @YCor said, in general for connected simply-connected compact groups, Out is the group of automorphisms of the corresponding Dynkin diagram. I don't think I understand your "$\mathrm{Sp}(2)$"; in your first comment you reference a $2\times 2$ matrix, but in your second you put it in $U(4)$. The second comment makes it seem that you do mean the group of type $B_2 = C_2$, in which case YCor is correct. Your first comment describes the "transpose" operation, which is not an automorphism in general (it is rather an "antiautomorphism": it reverse order of multiplication). But for $Sp$, ... – Theo Johnson-Freyd Jun 10 '15 at 00:02
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    ... depending on which basis you are using, the Lie algebra consists of transpose-invariant matrices. So your transpose map is probably nothing but the conjugation by the change-of-basis matrix that makes your matrices self-adjoint, and so is inner. – Theo Johnson-Freyd Jun 10 '15 at 00:05

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This is a compact semisimple Lie group whose Dynkin diagram is of type $B_2=C_2$ (one pair of vertices joined by an oriented double arrow), and has no nontrivial symmetry. So the outer automorphism group is trivial.

YCor
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