Out of curiosity:
Let $k$ be any field and $L$ be any Lie algebra over $k$ (of finite or infinite dimension).
Question: When does it happen that the group of Lie algebra automorphisms $\mathrm{Aut}(L)$ (that is: the $k$-linear bijections $\alpha: L \rightarrow L$ with $\alpha([x,y]) = [\alpha(x),\alpha(y)]$ for all $x,y \in L$), contains only the identity?
This is obviously the case when $L=0$, and if $L$ is the one dimensional Lie algebra over $k=\mathbb F_2$, and it feels as if these should be the only examples. But even though I can exclude a great many others, I do not have a full proof.
No other abelian $L$ qualifies, since for them $\mathrm{Aut}(L)$ is just the group of vector space automorphisms $GL(L)$, which has cardinality $\ge 2$ unless $\dim(L) =0$, or $\dim(L) =1$ and $\lvert k \rvert = 2$.
If $L \simeq L_1 \oplus L_2$ as Lie algebras, and $\lvert \mathrm{Aut}(L_1) \rvert \ge 2$, then $\lvert \mathrm{Aut}(L) \vert \ge 2$.
If $L$ contains a non-central nilpotent element $x$, and $char(k)$ is either $0$ or $\ge$ the nilpotency index of $\mathrm{ad}_x$ (i.e. the smallest $n$ such that $\mathrm{ad_x}^{\circ n} := \underbrace{\mathrm{ad}_x \circ ... \circ \mathrm{ad}_x}_{n \text{ times}} =0$), then $\mathrm{exp(ad}_x):= \displaystyle \sum_{n=0}^\infty \dfrac{\mathrm{ad_x}^{\circ n}}{n!}$ gives a non-trivial element of $\mathrm{Aut}(L)$. This at least suffices for finite-dimensional $L$ over algebraically closed fields of characteristic $0$ (see 1 -- note that YCor's answer gives a somewhat deeper insight into the relation between automorphisms and existence of $ad$-nilpotent elements). Can one improve this?
On matrix algebras $\mathfrak{gl}_n(k)$, we always have the automorphisms $X \mapsto -X^{tr}$, as well as $PGL_n(k)$ acting via conjugation. So to exclude automorphisms coming from that, we need that every faithful finite dimensional representation of $L$ is either not stable, or trivial, under each of these actions. That excludes every other example I can think of, but does not unfold to a full proof.