Decomposition of $sl_2(\mathbb{C})$ Let $L = sl_2(\mathbb{C})$. We call $L = H_0 \oplus H_1 \oplus H_2$
 an $orthogonal \ decomposition$ if  all components $H_i$ of decomposition are Cartan subalgebras and pairwise orthogonal with respect to the Killing form.
Now, suppose that $H_0$ is a Cartan subalgebra consisting of the diagonal matrices and $H_1$ is another Cartan subalgebra orthogonal to $H_0$ via the Killing Form. One can check easily that $H_1$ has a basis of the form:
 $H_1 = \Big < \begin{pmatrix}
0 & 1 \\
a & 0
\end{pmatrix} \Big >_\mathbb{C}$ for some $a \neq 0$. Thus, 
$L = H_0 \oplus \Big < \begin{pmatrix}
0 & 1 \\
a & 0
\end{pmatrix} \Big >_\mathbb{C}\oplus \Big < \begin{pmatrix}
0 & 1 \\
b & 0
\end{pmatrix} \Big >_\mathbb{C}$ for some $a, b \neq 0$.
Can we prove that either $a$ or $b$ (from above) must be $1$ in order to have the orthogonal decomposition? Or they don't have to be. If the answer is negative, can we still show that for each $a \neq 0$, there exists an automorphism $\phi \in Aut(L)$ such that $\phi(H_0) = H_0$ and 
$\phi(\begin{pmatrix}
0 & 1 \\
a & 0
\end{pmatrix}) = \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}$?
Note that the Killing form can be given by $K(A, B) = 4Tr(AB)$ for all $A, B \in sl_2(\mathbb{C})$.
 A: The orthogonal complement of $H_0$ in $\mathfrak sl_2$ with respect to the killing form is 
$$\{\begin{pmatrix} 0 & a \\  b & 0 \end{pmatrix} : a, b \in \mathbb{C} \}$$
and as you mentioned if you want to decompose the two dimensional space $H_0^{\perp}$ into a direct sum of Cartan subalgebras $H_1 \oplus H_2$, then you're forced to have $H_1 = \langle \begin{pmatrix} 0 & 1 \\ a & 0 \end{pmatrix} \rangle, H_2 = \langle \begin{pmatrix} 0 & 1 \\ b & 0 \end{pmatrix} \rangle$ for distinct, nonzero $a, b$.
If you want $H_1, H_2$ to be orthogonal, then you just need the trace $\begin{pmatrix} 0 & 1 \\ a & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ b & 0 \end{pmatrix} = \begin{pmatrix}b & 0 \\ 0 & a \end{pmatrix}$ to be zero.  So neither $a$ nor $b$ has to be one; you just need $a = -b$.  
For your question about the automorphism, I think the answer is no.  I believe every Lie algebra automorphism of $\mathfrak{sl}_2$ is the differential of an automorphism of algebraic groups of $\textrm{SL}_2$.  Every automorphism $\phi$ of $\textrm{SL}_2$ is know to be inner, say $x \mapsto gxg^{-1}$ for some $g \in \textrm{SL}_2$.  The differential $d\phi$ of such an automorphism is given by the same formula: $X \mapsto gXg^{-1}$.  
In order for $d\phi$ to fix $H_0$, $\phi$ has to fix the standard torus in $\textrm{SL}_2$, which means $g$ needs to be in that torus: so $g = \begin{pmatrix} c \\ & \frac{1}{c} \end{pmatrix}$.  From here you can see that what you want is impossible.
