Let us consider the total automorphism of special unitary group $SU(N)$ for $N>2$, we know that
$$
0 \to \text{Inn}(G)  \to \text{Aut}(G) \to \text{Out}(G) \to 0.
$$ 


-- For $G=SU(N)$ with  $N > 2$, 
we have

$\text{Z}(SU(N)) =\mathbb Z_{N}$,

$\text{Inn}(SU(N)) = PSU(N)$, 

and

$\text{Out}(SU(N)) = \mathbb Z_2$

>My question is that 
$$\text{Aut}(SU(N))=PSU(N) \times \mathbb Z_2?
$$
always true? Or does this answer alter when $N$ is odd or $N$ is 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
$$
B\text{Inn}(G)  \to B\text{Aut}(G) \to B\text{Out}(G) \to B^2\text{Inn}(G)  \to
$$
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?

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

$\text{Z}(SU(2)) =\mathbb Z_2$,

$\text{Inn}(SU(2)) = SO(3)$, 

$\text{Aut}(SU(2))=SO(3)$,

 and
$\text{Out}(SU(2)) = 0$. 

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Comments/hints are welcome. Answer will be awarded very soon. Thanks~!!!