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David Stewart
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Let me give this a try:

It seems you are really asking about whether there are any outer derivations of $Lie(G')$ in $Lie(G)$. [EDIT] This does not, unfortunately amount to exactly the question of whether outer derivations exist, because they may not be realised inside the normaliser of $Lie(G')$ in $Lie(G)$ (a good example is to take $G'$ is a torus; then the normaliser in $\mathfrak{gl}(Lie(G'))$ is itself as $G'$ is then a maximal torus of $GL(Lie(G'))$). Of course there are no outer derivations in characteristic $0$ for the Lie algebras of semisimple groups. In positive characteristic the situation is rather different. For classical simple Lie algebras (i.e. ones coming from algebraic groups---there are plenty more simple Lie algebras in char $p$ like the Witt algebras $W(n;\underline{m})$), there are no outer derivations in very good characteristic because the Killing form is non-degenerate. (You could read Seligman Modular Lie Algebras, p112.) So if you can reduce to this case then I suppose you're alright.

Now the Killing form is degenerate for $\mathfrak{psl}_{lp^r}$ (or indeed $\mathfrak{sl}_{lp^r}$) but you are ruling out these on the basis you are looking at the Lie algebra of an adjoint group and verily $\mathfrak{pgl}_{lp^r}$ has no outer derivations. However! There will possibly be an issue for $G_2$ in characteristic $2$ since the Lie algebra is isomorphic (qua Lie algebra) to $\mathfrak{psl}_4$ (amazing but true--e.g. count dimensions). Then this will have extra derivations and I would think this may give rise to a counterexample. Possibly also $F_4$ in characteristic $3$, $E_8$ in characteristic $5$, $\dots$ though I would take a small bet in favour of there are none except for $G_2$ in exceptional type. I would not be surprised to learn that there is a paper where someone has established exactly the outer derivations in all characteristics and for all isogeny types, but unfortunately I can't point you to one.

[EDIT: I just computed the $G_2$, $p=2$ example in GAP and tells me that it is its own normaliser. So maybe the answer two your question is always yes. At least for the exceptionals you could check this computationally. But there ought to be a more conceptual reason.]

David Stewart
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