Let $\mathfrak{g}$ be a Lie algebra. Then its automorphism group $Aut(\mathfrak{g})$ is a Lie group, and hence we may take its Lie algebra $Lie(Aut(\mathfrak{g}))$.
I'd like to say that this is equal to the Lie algebra of derivations $Der(\mathfrak{g})$. Is this true? Where can I find a reference?
On
$\DeclareMathOperator{\Aut}{Aut}$ It is in fact true that $\text{Lie}(\Aut(L))=\text{Der}(L)$, as you say. There are a lot of questions on Stack Exchange asking about this, and how to prove it. But no answers seem to give a proof.
[Additionally, the only source I saw out of all of these questions, was the alternative answer to this particular question. Unfortunately, the source given is just showing it is true for the real numbers assuming it is true for the complex numbers. The source gives its own source for the proof for the complex number case, but the limited access I had on Google Books did not let me see the references.]
Luckily, I found that this is an exercise in Fulton and Harris's Representation Theory, specifically $8.27$, and it is one of the few exercises without a hint in the back, so it cannot be that hard.
And it actually is not. Let $G=\{g\in \text{GL}(L)| g[x,y]=[gx,gy], x,y\in L\}$. Then, in order to get the tangent space of $G$ at $1$, i.e. the Lie algebra of $G$, we just have to differentiate the equation $g[x,y]=[gx,gy]$ and plug in $g=1$. If we differentiate at $g$ in the direction of $\delta$, we get $\delta([x,y])=[gx,\delta(y)]+[\delta(x),gy]$, and so plugging in $g=1$ gives us $\delta([x,y])=[x,\delta(y)]+[\delta(x),y]$. Thus, $\delta$ is in the tangent space of $1$ if and only if $\delta$ is a derivation on $L$. (And the Lie products are the same, because they are both given by the commutator $[\delta_1,\delta_2]=\delta_1\delta_2-\delta_2\delta_1$).
This is a standard fact from the theory of Lie algebras. There are several references, say over $\Bbb{R}$ and $\Bbb{C}$, e.g., Proposition 1.25 in the book The Structure of Complex Lie Groups.