# What is the no. of automorphism on U(6) [closed]

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?

## closed as off-topic by YuiTo Cheng, Shailesh, Hayk, Thomas Shelby, LeucippusMay 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.
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!