-2
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I tried with U(6)~U(2)×U(3)~Z1×Z2~Z2 and hence aut(U(6)) is isomorphic to Z2. Is it the right way?

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closed as off-topic by YuiTo Cheng, Shailesh, Hayk, Thomas Shelby, Leucippus May 22 at 7:26

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$U(6) =\{1,5 \} \simeq \mathbb{Z}_2$.

Now $Aut(\mathbb{Z}_n)=U(n)$ and hence $ Aut((U(6))=Aut(\mathbb{Z}_2)=U(2)=\{1\}$.

Thus only $1$ automorphism is there.

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0
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There are only two elements in $U(6)$ and note that automorphism maps identity to identity, we map $1$ into itself. Since this map is in particular one to one, so we must map the other element to itself. Hence there is actually only one automorphism!

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