Let $p:\tilde{G}\to G$ be a universal cover. Then who are the outer automorphisms $\operatorname{out}(G)$ in terms of $\operatorname{out}(\tilde{G})$? For example, let $p:\operatorname{SU}(3)\to G$ be a universal cover. May I say that the outer automorphisms of $G$ are of the form $p(g)\mapsto p(\overline{g})$, since the outer automorphisms of $\operatorname{Out}\operatorname{SU}(3)=\langle \sigma\rangle$, sigma being the complex conjugation?
In general, is there any simple way to write $\operatorname{Out}(G)$ in terms of $\operatorname{Out}(\tilde{G})$?
We have that if $\varphi\in\operatorname{Out}(G)$, as $\tilde{G}$ is simply connected, $p\circ\varphi$ lifts to a homomorphism $\tilde{\varphi}:\tilde{G}\to\tilde{G}$. But $\ker \tilde{\varphi}=p^{-1}(1)$. Is there any way to turn this $\tilde{\varphi}$ into an automorphism? In my specific case I may assume that $G=\tilde{G}/Z(\tilde{G})$.
 A: Let me assume $G$ is connected so that all the lifts make sense and are uniquely defined and group morphisms.
$\ker\tilde\varphi $ is not $p^{-1}(1)$ ! In fact $\tilde\varphi$ is always an automorphism.
Indeed, you can also lift $\psi = \varphi^{-1}$ to $\tilde \psi$ and $\tilde \psi\tilde\varphi$ is a lift of $id_G$ which sends $1$ to $1$ so it must be $id_{\tilde G}$, and similarly for $\tilde\varphi\tilde \psi$.
In fact, this clearly provides a group monomorphism $Aut(G)\to Aut(\tilde G)$. Now if $\varphi$ is inner, say $x\mapsto gxg^{-1}$, and if $g_0$ is a lift of $g$, then $y\mapsto g_0yg_0^{-1}$ is a lift of of $\varphi$ (which in fact doesn't depend on the choice of $g_0$), so it factors as a morphism $Out(G)\to Out(\tilde G)$.
In fact, because $\ker p$ is central in $\tilde G$, $Inn(G)\to Inn(\tilde G)$ is an isomorphism, so this map on outer automorphism groups is also injective.
In the specific case you describe, $G = \tilde G/Z(\tilde G)$, since $Z(\tilde G)$ is not only normal but characteristic, that is, it's preserved by arbitrary automorphisms, it follows that in fact, $Aut(G)\to Aut(\tilde G)$ is an isomorphism.
More generally, I'm not sure what can be said; e.g. if we look at $\mathbb R\to S^1$, then  $Aut(\mathbb R) = \mathbb R^\times$ (I'm thinking in terms of continuous automorphisms for everything to make sense of course), whereas $Aut(S^1) = C_2$, the injection being given by $-1\mapsto -1$ (if we view $C_2$ as $\{\pm 1\}$)
