Automorphism group of the projective unitary group PU(N) and SO(N)

I think most of the answer is in your MathOverflow post, which you should link.

Any automorphism of a Lie group $G$ induces a Lie algebra automorphism on the Lie algebra $\mathfrak g$. Moreover, a compact Lie group $G$ is generated by the image of a small neighborhood of the identity in $\mathfrak g$ under the exponential map, so that the tangent map $$\mathrm{Aut }\, G \to \mathrm{Aut}\, \mathfrak g$$ is injective. In fact, if $G$ is simply-connected, then this map is bijective, as mentioned in the comments to your MO question. I believe that you already know the automorphisms of the Lie algebra, so if $\pi\colon\tilde G \to G$ is the universal cover, you know $\mathrm{Aut}\,\tilde G$ as well.

From the injectivity of the tangent map, you thus expect to be able to identify $\mathrm{Aut }\, G$ with a subgroup of $\mathrm{Aut}\, \tilde G$ that has the same tangent information (and so, particularly, looks the same near the identity, $\pi$ being a finite-sheeted covering. If we write $K = \ker \pi$, so that $G = \tilde G/K$, then a (continuous) automorphism $\tilde \phi$ of $\tilde G$ will induce a well-defined automorphism $\phi$ of $G$ by $\phi(gK) := \tilde\phi(g)K$ if and only if $\tilde\phi(K) = K$. (Moreover, since $\pi$ is a local bijection near $1 \in \tilde G$, this is the only option.)

So assuming you understand $\mathrm{Aut}\, \mathfrak g \cong \mathrm{Aut}\, \tilde G$, what remains is to identify $\ker(\tilde G \to G)$ and to determine which automorphisms preserve it.

In the case of the special unitary group, the kernel is the center of $\mathrm{SU}(N)$ (which is the subgroup of diagonal matrices $\zeta \cdot I$ for $\zeta$ an $N^{\mathrm{th}}$ root of unity). But the center of a group is preserved by any automorphism.

The corresponding question for $\mathrm{SO}(2N)$ may be more interesting because the kernel of the covering map $\mathrm{Spin}(2N) \to \mathrm{SO}(2N)$ is not the whole center and so there is something left to check.