Let us consider the automorphism group of the special unitary group $G=SU(N)$. 


We know there is an exact sequence:
$$
0 \to \text{Inn}(G) \to \text{Aut}(G) \to \text{Out}(G) \to 0.
$$ 


For $G=SU(2)$, we have:


* $\text{Z}(SU(2)) =\mathbb Z\_2$,
* $\text{Inn}(SU(2)) = SO(3)$,
* $\text{Out}(SU(2)) = 0$,


And so $\text{Aut}(SU(2))=SO(3)$.


For $N > 2$, we have:


* $\text{Z}(SU(N)) =\mathbb Z\_{N}$,
* $\text{Inn}(SU(N)) = PSU(N)$,
* $\text{Out}(SU(N)) = \mathbb Z\_2$.


My question is:



> 
> Does $\text{Aut}(SU(N))=PSU(N) \times \mathbb Z\_2$?
>  If not, does this answer depend on whether $N$ is odd or even?
> 
> 
> 


It looks to me that there is a nontrivial fibration depending on something like $H^2(B\mathbb Z\_2,PSU(N))$ due to $$ B\text{Inn}(G) \to B\text{Aut}(G) \to B\text{Out}(G) \to B^2\text{Inn}(G) \to$$ and
thus
$$ BPSU(N) \to B\text{Aut}(G) \to B\mathbb Z\_2 \to B^2PSU(N) \to$$


But I do not know how to define $H^2(B\mathbb Z\_2,PSU(N))$, if this is a correct thing to ponder.



Automorphism group of the special unitary group $SU(N)$
Let $G$ be a compact simple simply connected Lie group. Then any automorphism of $G$ determines an automorphism of its Lie algebra $\mathfrak{g}$ and visa versa. So $\mathrm{Aut}(G)$ is naturally isomorphic to the linear group $\mathrm{Aut}(\mathfrak{g})$.


The sequence $1\to \mathrm{Inn}(\mathfrak{g})\to \mathrm{Aut}(\mathfrak{g})\to \mathrm{Out}(\mathfrak{g})\to 1$ is [split](https://mathoverflow.net/q/14735/12218).


Moreover, $\mathrm{Out}(G)\cong\mathrm{Out}(\mathfrak{g})\cong \mathrm{Aut}(D\_\mathfrak{g})$ where $D\_\mathfrak{g}$ is the Dynkin diagram of $\mathfrak{g}$.


The upshot is:


$$\mathrm{Aut}(G)\cong \mathrm{Aut}(\mathfrak{g})\cong \mathrm{Inn}(\mathfrak{g})\rtimes \mathrm{Out}(G)\cong \mathrm{Inn}(\mathfrak{g})\rtimes \mathrm{Aut}(D\_\mathfrak{g}).$$


For types $A\_1, B\_n, C\_n, G\_2, F\_4, E\_7, E\_8$ there are no symmetries of the Dynkin diagram. For $A\_n$ ($n>1$), $D\_n$ ($n\not=4$), and $E\_6$, we have $\mathrm{Aut}(D\_\mathfrak{g})\cong \mathbb{Z}/2\mathbb{Z}$. And in the final case of $D\_4$, the symmetry group is the symmetric group on three letters.


In particular, as stated in the comments: $$\mathrm{Aut}(\mathrm{SU}(n))\cong\left\{\begin{array}{ll}\mathrm{PSU}(n)\rtimes \mathbb{Z}/2\mathbb{Z},&\text{ if } n\geq 3\\ \mathrm{PU}(2),&\text{ if }n=2. \end{array}\right.$$


