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.