Are there 'authentic' outer derivations of Lie algebras? Let $\mathfrak g$ be a finite-dimensional Lie algebra and let $\mathfrak g\subset\mathfrak h$ be an extension of $\mathfrak g$. Then every derivation of $\mathfrak h$ induces a derivation of $\mathfrak g$ by restriction. In particular, every inner derivation of $\mathfrak h$ induces a derivation of $\mathfrak g$, which is not necessarily inner.
More precisely, there is an exact sequence
$$0\rightarrow Z(\mathfrak h)\rightarrow{\rm Ann}_{\mathfrak h}(\mathfrak g)\rightarrow{\rm
Inn}(\mathfrak h)\rightarrow{\rm Der}(\mathfrak g),$$
which simply means that an element $h\in\mathfrak h$ induces an inner derivation (possibly trivial) of $\mathfrak g$ if and only if $h+h'\in\mathfrak g$ for some $h'\in{\rm Ann}_{\mathfrak h}(\mathfrak g)$. Clearly, every inner derivation of $\mathfrak g$ is induced by an inner derivation of $\mathfrak h$, since $\mathfrak g\subset\mathfrak h$.
Now, let ${\rm Out}_{\mathfrak h}(\mathfrak g)$ be defined as the vector space that completes the above exact sequence:
$$0\rightarrow Z(\mathfrak h)\rightarrow{\rm Ann}_{\mathfrak h}(\mathfrak g)\rightarrow{\rm
Inn}(\mathfrak h)\rightarrow{\rm Der}(\mathfrak g)\rightarrow{\rm Out}_{\mathfrak h}(\mathfrak
g)\rightarrow0,$$
i.e., ${\rm Out}_{\mathfrak h}(\mathfrak g)$ consists of the outer derivations of $\mathfrak g$ modulo those that are induced by inner derivations of $\mathfrak h$.

Given a Lie algebra $\mathfrak g$, is there an extension $\mathfrak g\subset\mathfrak h$ such that ${\rm Out}_{\mathfrak h}(\mathfrak g)=0$?

In other words, is it always possible to extend a Lie algebra $\mathfrak g$ in such a way that all its derivations are induced by inner derivations of the extension?
In order to get some insight about how ${\rm Out}_{\mathfrak h}(\mathfrak g)$ behaves, I tried to consider some particular cases. For example, if $\mathfrak g$ has no outer derivations, then obviously ${\rm Out}_{\mathfrak h}(\mathfrak g)={\rm Out}(\mathfrak g)=0$ for any extension $\mathfrak g\subset\mathfrak h$.
On the other hand, if $\mathfrak g\subset\mathfrak h$ is a central extension, i.e., $\mathfrak g\subset Z(\mathfrak h)$ or equivalently ${\rm Ann}_{\mathfrak h}(\mathfrak g)=\mathfrak h$, then it's easy to see that ${\rm Der}_{\mathfrak h}(\mathfrak g)={\rm Der}(\mathfrak g)=\mathfrak{gl}(\mathfrak g)$. However, if $\mathfrak g\subset\mathfrak h$ is just an abelian extension, then this equality doesn't necessarily hold anymore.
For instance, let $\mathfrak h=\mathfrak r_{3,\lambda}(\Bbb C)$ be the $3$-dimensional complex Lie algebra defined by the relations $[e_1,e_2]=e_2$, $[e_1,e_3]=\lambda e_3$, where $\lambda\in\Bbb C$, $0<|\lambda|\le1$, and let $\mathfrak g\subset\mathfrak h$ be the ideal generated by $\{e_2,e_3\}$. Then $\mathfrak g\subset\mathfrak h$ is clearly an abelian extension.
A simple computation shows that $\dim{\rm Out}(\mathfrak g)=\dim\mathfrak{gl}(\mathfrak g)=4$, but ${\rm Out}_{\mathfrak h}(\mathfrak g)=3$. Indeed, the element $e_1\in\mathfrak h$ induces an outer derivation of $\mathfrak g$ given by the following matrix in terms of the basis $\{e_2,e_3\}$:
$$\pmatrix{
1&0\\
0&\lambda}$$
Also, this example helps to understand that ${\rm Out}_{\mathfrak h}(\mathfrak g)$ is not necessarily a Lie algebra. Indeed, if $\lambda\ne1$, then the vector subspace generated by the above matrix isn't an ideal in $\mathfrak{gl}(\mathfrak g)$; it's normalizer consists of diagonal matrices.
If $\lambda=1$, nevertheless, then ${\rm Out}_{\mathfrak h}(\mathfrak g)$ is a Lie algebra.

Is there a special name for Lie algebra extensions such that ${\rm Out}_{\mathfrak h}(\mathfrak g)$ is a Lie algebra?

Besides answers to the above questions, any references related to these matters will be really appreciated.
 A: I’m not sure about the second question, but I think the first question can be answered as “Yes” with the help of semidirect products.
Let $\mathfrak{g}$ and $\mathfrak{k}$ be two Lie algebras and let $\theta$ be a homomorphism of Lie algebras from $\mathfrak{k}$ to ${\rm Der}(\mathfrak{g})$.
We can then form the semidirect product $\mathfrak{k} \ltimes_\theta \mathfrak{g} =: \mathfrak{h}$.
Its underlying vector space is given by the direct sum $\mathfrak{k} \oplus \mathfrak{g}$, and its Lie bracket is given by
$$
  [ (x_1, y_1), (x_2, y_2) ]
  =
  ( [x_1, x_2], \theta(x_1)(y_2) - \theta(x_2)(y_1) + [y_1,y_2] ) \,.
$$
The inclusion $i$ from $\mathfrak{k}$ to $\mathfrak{h}$ given by $i(x) = (x,0)$ is an injective homomorphism of Lie algebras, which identifies $\mathfrak{k}$ with a Lie subalgebra of $\mathfrak{h}$.
The inclusion $j$ from $\mathfrak{g}$ to $\mathfrak{h}$ given by $j(y) = (0,y)$ is also an injective homomorphism of Lie algebras, but it identifies $\mathfrak{g}$ with a Lie ideal of $\mathfrak{h}$.
The Lie bracket on $\mathfrak{h}$ is built in precisely such a way that
\begin{align*}
  [i(x), j(y)]
  &=
  [ (x,0), (0,y) ]
  \\
  &=
  ( [x,0], \theta(x)(y) - \theta(0)(0) + [0,y] )
  \\
  &=
  (0, \theta(x)(y))
  \\
  &=
  j( \theta(x)(y) )
\end{align*}
for all $x \in \mathfrak{k}$, $y \in \mathfrak{g}$.
In other words, the restriction of the inner dervation $[i(x), -]$ to $\mathfrak{g}$ (when regarded a Lie ideal of $\mathfrak{h}$) is the derivation $\theta(x)$.
This constructions allows us to turn arbitrary derivations of $\mathfrak{g}$ into inner derivations on a suitable extension $\mathfrak{h}$ of $\mathfrak{g}$.
An extreme case of this is $\mathfrak{k} = \operatorname{Der}(\mathfrak{g})$ and $\theta = \mathrm{id}$.
The resulting semidirect product
$$
  \mathfrak{h} := \operatorname{Der}(\mathfrak{g}) \ltimes_{\mathrm{id}} \mathfrak{g}
$$
is an extension of $\mathfrak{g}$, and every derivation of $\mathfrak{g}$ comes from an inner derivation of $\mathfrak{h}$.
More precisely, if $\delta$ is a derivation of $\mathfrak{g}$ then the restriction of the inner derivation $[(\delta, 0), -]$ of $\mathfrak{h}$ to $\mathfrak{g}$ is precisely $\delta$.
