Automorphism group of Lie algebra $\mathfrak{g\oplus g}$ Let $\mathfrak g$ be the Lie algebra of strictly upper triangular 3x3 matrices. How can I determine the group $\operatorname{Aut}(\mathfrak{g\oplus g},\Delta\mathfrak g)$, where $\Delta\colon\mathfrak g\to\mathfrak {g\oplus g}$ is the diagonal embedding?
By $\operatorname{Aut}(\mathfrak{g\oplus g},\Delta\mathfrak g)$ I mean the automorphisms $\phi\in\operatorname{Aut}(\mathfrak{g\oplus g})$ such that $\phi(\Delta\mathfrak g)\subset\Delta\mathfrak g$.
 A: I really don't know what you mean by how can I find this group: you can simply compute it!
The algebra $\def\g{\mathfrak g}\g$ has a basis $\{x,y,z\}$ such that $z$ is central and $[x,y]=z$. It follows that $\g\oplus g$ has a basis $\{x_1,y_1,z_1,x_2,y_2,z_2\}$ with $[x_i,y_i]=z_i$ for $i\in\{1,2\}$  and all other brackets between elements of the basis equal to zero. 
Let $f:\g\oplus\g\to\g\oplus\g$ be an automorphism.
The derived algebra $\def\h{\mathfrak h}\h=[\g,\g]$ is spanned by $\{z_1,z_2\}$ and coincides with the center of $\g$. Since this is obviously preserved by every automorphism, there are scalars $a_i$, $b_1$ such that $f(z_i)=a_iz_1+b_iz_2$. Since the restriction of $f$ to a map $\h\to\h$ must be invertible, we must have $\det\begin{pmatrix}a_1&a_2\\b_1&b_2\end{pmatrix}\neq0$.
There are scalars $p_i$, $q_i$, $r_i$, $s_i$, $t_i$, $u_i$, $v_i$, $w_i$ and elements $\xi_i$, $\zeta_i\in\h$ such that $f(x_i)=p_ix_1+q_iy_1+r_ix_2+s_iy_2+\xi_i$ and $f(y_i)=t_ix_1+u_iy_1+v_ix_2+w_iy_2+\zeta_i$. It follows that $[f(x_i),f(y_i)]=(p_iu_i-q_it_i)z_1+(r_iw_i-s_iv_i)z_2$, and this must be equal to $f([x_i,y_i])=f(z_i)=a_iz_1+a_iz_2$. It follows that we must have $a_i=p_iu_i-q_it_i$ and $b_i=r_iw_i-s_iv_i$. Since $[x_1,x_2]=[x_1,y_2]=[x_2,y_1]=[y_1,y_2]=0$, we must have 
\begin{align}
&[f(x_1),f(x_2)]=(p_1q_2-q_1p_2)z_1+(r_1s_2-s_1r_2)z_2=0,\\
&[f(x_1),f(y_2)]=(p_1u_2-q_1t_2)z_1+(r_1w_2-s_1v_2)z_2=0,\\
&[f(x_2),f(y_1)]=(p_2u_1-q_2t_1)z_1+(r_2w_1-s_2v_1)z_2=0,\\
&[f(y_1),f(y_2)]=(t_1u_2-u_1t_2)z_1+(v_1w_2-w_1v_2)z_2=0,
\end{align}
and this means that 
\begin{align}
p_1q_2-q_1p_2=r_1s_2-s_1r_2=0,\\
p_1u_2-q_1t_2=r_1w_2-s_1v_2=0,\\
p_2u_1-q_2t_1=r_2w_1-s_2v_1=0,\\
t_1u_2-u_1t_2=v_1w_2-w_1v_2=0,
\end{align}
So far we have expressed all the conditions that express the fact that $f$ is an isomorphism. If you want $f$ to preserve $\Delta f$, then this gives you more conditions. Write them out. Then look at what you got.
I won't write the details, because this is very boring and you are, after all, who is interested in the result! :-)
