Every automorphism of $\mathfrak{sl}(2,\mathbb{C})$ is given by conjugation with some $u \in \mathrm{SL}(2,\mathbb{C})$ Let  $\mathfrak g = \mathfrak{sl}(2, \mathbb{C})$. Let $\gamma \in \operatorname{Aut}(\mathfrak{g})$. How to show that $\gamma$ is conjugation by some $u \in \mathrm{SL}(2, \mathbb C)$?
 A: The following uses the theory of Jordan decomposition in finite dimensional semisimple complex Lie algebras (namely in $\mathfrak{sl}_2(\mathbb{C})$) and some basic knowledge about the finite dimensional representations of $\mathfrak{sl}_2(\mathbb{C})$. There is probably a much shorter and easier way to do this, but it’s the best I’m coming up with right now.

Let $(e,h,f)$ be the usual standard basis of $\mathfrak{g} := \mathfrak{sl}_2(\mathbb{C})$, i.e.
$$
 e =
 \begin{pmatrix}
  0 & 1 \\
  0 & 0
 \end{pmatrix}, \quad
 h =
 \begin{pmatrix}
  1 & \phantom{-}0 \\
  0 & -1
 \end{pmatrix}, \quad
 f =
 \begin{pmatrix}
  0 & 0 \\
  1 & 0
 \end{pmatrix}.
$$
Because $h$ is diagonalizable as a matrix it is a semisimple element of $\mathfrak{g}$. As $\gamma$ is a Lie algebra automorphism we find that $\gamma(h)$ is also a semisimple element of $\mathfrak{g}$.
Because $\gamma(h)$ is a semisimple of $\mathfrak{g}$ it follows that $\gamma(h)$ is diagonalizable as a matrix, with eigenvalues $\lambda$ and $-\lambda$ (because $\gamma(h)$ is traceless). We actually have $\{\lambda, -\lambda\} = \{1,-1\}$:
The natural representation $\rho \colon \mathfrak{g} \to \mathfrak{gl}(\mathbb{C}^2)$ is an irreducible representation of $\mathfrak{sl}_2(\mathbb{C})$. As $\gamma \colon \mathfrak{g} \to \mathfrak{g}$ is an Lie algebra automorphism it follows that the compositon $\rho \circ \gamma \colon \mathfrak{g} \to \mathfrak{gl}(\mathbb{C}^2)$ is also an irreducible representation (both $\rho$ and $\rho \circ \gamma$ have precisely the same subrepresentations). By the usual classification of finite dimensional irreducible representations of $\mathfrak{sl}_2(\mathbb{C})$ we find that $\mathbb{C}^2$ decomposes into the $h$-weight spaces $1$ and $-1$ with respect to $\rho \circ \gamma$. But these weight spaces are precisely the eigenspaces of $\gamma(h)$, and the weights $1$ and $-1$ are precisely the eigenvalues of $\gamma(h)$.

As $\gamma(h)$ is diagonalizable with eigenvalues $1$ and $-1$ there exists some $u \in \mathrm{GL}_2(\mathbb{C})$ with
$$
 u \gamma(h) u^{-1} =
 \begin{pmatrix}
  1 & \phantom{-}0  \\
  0 & -1
 \end{pmatrix}
 = h.
$$
We can assume that $u \in \mathrm{SL}_2(\mathbb{C})$, i.e. $\det u = 1$. Otherwise we could choose some $\mu \in \mathbb{C} \setminus \{0\}$ with $\mu^2 = \det u$ and replace $u$ with $\mu^{-1} u$.
Now consider the Lie algebra automorphism $\beta \colon \mathfrak{g} \to \mathfrak{g}$ with $\beta(x) = u x u^{-1}$. We already have $\beta(\gamma(h)) = h$. It remains to show the following:

Lemma: Let $\alpha \colon \mathfrak{g} \to \mathfrak{g}$ be a Lie algebra automorphism with $\alpha(h) = h$. Then there exists $v \in \mathrm{SL}_2(\mathbb{C})$ with $\alpha(x) = v x v^{-1}$ for all $x \in \mathfrak{g}$.

From this it then follows that $\beta(\gamma(x)) = vxv^{-1}$, and thus
$$
 \gamma(x)
 = \beta^{-1}(vxv^{-1})
 = u^{-1} v x v^{-1} u
 = (u^{-1} v) x (u^{-1} v)^{-1}
$$
with $u^{-1} v \in \mathrm{SL}_2(\mathbb{C})$.

To prove the lemma notice that
$$
 [h,\alpha(e)]
 = [\alpha(h),\alpha(e)]
 = \alpha([h,e])
 = 2\alpha(e),
$$
so $\alpha(e)$ is an eigenvalues of $\operatorname{ad}(h)$ with eigenvalue $2$. But the corresponding eigenspace is spanned by $e$, so $\alpha(e) = \lambda e$ for some $\lambda \in \mathbb{C}$. Similarly $\alpha(f) = \lambda' f$ for some $\lambda' \in \mathbb{C}$. Because
$$
 h
 = \alpha(h)
 = \alpha([e,f])
 = [\alpha(e), \alpha(f)]
 = \lambda \lambda' [e,f]
 = \lambda \lambda' h
$$
we have $\lambda \lambda' = 1$, so $\lambda$ is non-zero with $\lambda' = \lambda^{-1}$. So $\alpha$ is given by $\alpha(e) = \lambda e$, $\alpha(h) = h$ and $\alpha(f) = \lambda^{-1} f$. Choosing $\mu \in \mathbb{C}$ with $\mu^2 = \lambda$ set
$$
 v =
 \begin{pmatrix}
  \mu & 0       \\
  0   & \mu^{-1}
 \end{pmatrix}.
$$
Then a straightforward calculation shows that $v e v ^{-1} = \lambda e$, $v h v^{-1} = h$ and $v f v^{-1} = \lambda^{-1} f$.
