Automorphism group of the special unitary group $SU(N)$ 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.
 A: 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.
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: $$\mathrm{Aut}(\mathrm{SU}(n))\cong\left\{\begin{array}{ll}\mathrm{PSU}(n)\rtimes \mathbb{Z}/2\mathbb{Z},&\text{ if } n\geq 3\\ \mathrm{PU}(2),&\text{ if }n=2. \end{array}\right.$$
