Embeddings of several Lie groups and their geometry embedding The question concerns some problems about the Lie groups and representations. And the geometry embedding of the several Lie groups.
We start from a fixed common special unitary group SU(2), with the given SU(2) representations, then try to generate two Lie groups SU(4) and SU(4)' out of the fixed SU(2). The question concerns: what is the minimal Lie group $$G =?$$ that contains both the two generated SU(4) and SU(4)'? (Here SU(4) and SU(4)' are isomorphic as the Lie group, but the ways that it acts on the SU(2) representations are different.)
See the figure for an imaginative illustration below:

Consider the SU(2) representations: the 1-dimensional ${\bf 1}$ and the 2-dimensional ${\bf 2}$. We take four of ${\bf 1}$ and three of ${\bf 2}$ of SU(2) in total. So they combine to be:
$$
{\bf 1} \oplus {\bf 1} \oplus {\bf 1} \oplus {\bf 1} \oplus {\bf 2} \oplus {\bf 2}\oplus {\bf 2} \text{ of SU(2)}.
$$
More precisely, let us distinguish them by some sublabel (subindices):
$$
{\bf 1}_a \oplus {\bf 1}_b \oplus {\bf 1}_{c'} \oplus {\bf 1}_{d'} \oplus {\bf 2}_{A'} \oplus {\bf 2}_{C} \oplus {\bf 2}_{D} \text{ of SU(2)}.
$$
1.First SU(4)
Now we generate the first SU(4) by defining a 4-dimensional ${\bf 4}$ and a 6-dimensional ${\bf 6}$, as
$$
{\bf 4} \text{ of SU(4)}\mapsto {\bf 1}_a \oplus {\bf 1}_b \oplus {\bf 2}_{A'}
\text{ of SU(2)}
$$
and
$$
{\bf 6} \text{ of SU(4)}\mapsto {\bf 1}_{c'} \oplus {\bf 1}_{d'} \oplus {\bf 2}_{C} \oplus {\bf 2}_{D} \text{ of SU(2)}
$$
So using the red and orange colors to break down the above information, we have
$$
\color{red}{{\bf 4}} \oplus
\color{orange}{\bf 6} \text{ of SU(4)}\mapsto \color{red}{({\bf 1}_a \oplus {\bf 1}_b \oplus {\bf 2}_{A'})} \oplus
\color{orange}{({\bf 1}_{c'} \oplus {\bf 1}_{d'} \oplus {\bf 2}_{C} \oplus {\bf 2}_{D})}
\text{ of SU(2)}
$$
Or the original ${\bf 1}_a \oplus {\bf 1}_b \oplus {\bf 1}_{c'} \oplus {\bf 1}_{d'} \oplus {\bf 2}_{A'} \oplus {\bf 2}_{C} \oplus {\bf 2}_{D}$ becomes
$$
\color{red}{{\bf 4}} \oplus
\color{orange}{\bf 6} \text{ of SU(4)}\mapsto
\color{red}{{\bf 1}_a} \oplus \color{red}{{\bf 1}_b} \oplus \color{orange}{{\bf 1}_{c'}} \oplus\color{orange}{ {\bf 1}_{d'}} \oplus \color{red}{{\bf 2}_{A'}} \oplus \color{orange}{{\bf 2}_{C}} \oplus \color{orange}{{\bf 2}_{D}}\text{ of SU(2)}. \tag{1}
$$
We can determine how the SU(4) is generated out of the chosen representations of SU(2).
2.Second SU(4)'
Now we generate the second SU(4)' by defining another 4-dimensional ${\bf 4}'$ and another 6-dimensional ${\bf 6}'$, as
$$
{\bf 4}' \text{ of SU(4)'}\mapsto {\bf 1}_{c'} \oplus {\bf 1}_{d'}  \oplus {\bf 2}_{A'}
\text{ of SU(2)}
$$
and
$$
{\bf 6}' \text{ of SU(4)'}\mapsto {\bf 1}_{a} \oplus {\bf 1}_{b} \oplus {\bf 2}_{C} \oplus {\bf 2}_{D} \text{ of SU(2)}
$$
So using the blue and purple color to break down the above information, we have
$$
\color{blue}{{\bf 4}} \oplus
\color{purple}{\bf 6} \text{ of SU(4)'}\mapsto \color{blue}{( {\bf 1}_{c'} \oplus {\bf 1}_{d'}  \oplus {\bf 2}_{A'})} \oplus
\color{purple}{({\bf 1}_{a} \oplus {\bf 1}_{b} \oplus {\bf 2}_{C} \oplus {\bf 2}_{D})}
\text{ of SU(2)}
$$
Or the original ${\bf 1}_a \oplus {\bf 1}_b \oplus {\bf 1}_{c'} \oplus {\bf 1}_{d'} \oplus {\bf 2}_{A'} \oplus {\bf 2}_{C} \oplus {\bf 2}_{D}$ becomes
$$
\color{blue}{{\bf 4}} \oplus
\color{purple}{\bf 6} \text{ of SU(4)'}\mapsto
\color{purple}{{\bf 1}_a} \oplus \color{purple}{{\bf 1}_b} \oplus \color{blue}{{\bf 1}_{c'}} \oplus\color{blue}{ {\bf 1}_{d'}} \oplus \color{blue}{{\bf 2}_{A'}} \oplus \color{purple}{{\bf 2}_{C}} \oplus \color{purple}{{\bf 2}_{D}} \text{ of SU(2)}. \tag{2}
$$
We can determine how the SU(4)' is generated out of the chosen representations of SU(2).
Question
Again, we should contrast:
$$
\color{red}{{\bf 4}} \oplus
\color{orange}{\bf 6} \text{ of SU(4)}\mapsto
\color{red}{{\bf 1}_a} \oplus \color{red}{{\bf 1}_b} \oplus \color{orange}{{\bf 1}_{c'}} \oplus\color{orange}{ {\bf 1}_{d'}} \oplus \color{red}{{\bf 2}_{A'}} \oplus \color{orange}{{\bf 2}_{C}} \oplus \color{orange}{{\bf 2}_{D}} \text{ of SU(2)}. \tag{1}
$$
$$
\color{blue}{{\bf 4}} \oplus
\color{purple}{\bf 6} \text{ of SU(4)'}\mapsto
\color{purple}{{\bf 1}_a} \oplus \color{purple}{{\bf 1}_b} \oplus \color{blue}{{\bf 1}_{c'}} \oplus\color{blue}{ {\bf 1}_{d'}} \oplus \color{blue}{{\bf 2}_{A'}} \oplus \color{purple}{{\bf 2}_{C}} \oplus \color{purple}{{\bf 2}_{D}} \text{ of SU(2)}. \tag{2}
$$
We start from a fixed common special unitary group SU(2), given the SU(2) representations, then try to generate two Lie groups SU(4) and SU(4)' out of the fixed SU(2). The two SU(4) and SU(4)' overlap with the same Lie group SU(2), in the sense of geometry embedding. The question concerns: what is the minimal Lie group, called $G$, that contains both the two generated SU(4) and SU(4)'? (Is that, for example, SU(6) or SU(8)?)
Again, the question: what is the minimal Lie group
$$G =?$$
See the figure for an imaginative illustration above.
For your answer, give a precise $G$, and what is the $G$ as a minimal manifold (minimal surface) satisfying all conditions?
 A: This may be a partial answer.

*

*For your setup, you can look at the branching rules
$$
su(4) ⊃ su(2) ⊕ su(2) ⊕ u(1)
$$
for the regular Lie subalgebra (R). Then you find the branching rule of
the representation ${\bf 4}$ of $su(4)$ decomposes as the representation of $su(2) ⊕ su(2) ⊕ u(1)$:
$$
{\bf 4} = ({\bf 2}, {\bf 1},1) ⊕ ({\bf 1},{\bf 2}, −1).
$$


*You can look at the branching rules
$$
su(6) ⊃ su(4) ⊕ su(2) ⊕ u(1)
$$
for the regular Lie subalgebra (R). Then you find the branching rule of
the representation ${\bf 6}$ of $su(6)$ decomposes as the representation of $su(4) ⊕ su(2) ⊕ u(1)$:
$$
{\bf 6} = ({\bf 4}, {\bf 1},1) ⊕ ({\bf 1},{\bf 2}, −2).
$$


*Then you see you two conditions $$
{\bf 4} \text{ of SU(4)}\mapsto {\bf 1}_a \oplus {\bf 1}_b \oplus {\bf 2}_{A'}
\text{ of SU(2)}
$$
$$
{\bf 4}' \text{ of SU(4)'}\mapsto {\bf 1}_{c'} \oplus {\bf 1}_{d'}  \oplus {\bf 2}_{A'}
\text{ of SU(2)}
$$
can be combined to be a single representation ${\bf 6}$ of $su(6)$, such that under the first decomposition:
$$
su(6) ⊃ su(4) ⊕ su(2) ⊕ u(1)⊃ su(2) ⊕ su(2)⊕ su(2) ⊕ u(1)⊕ u(1)
$$
such that
$$
{\bf 6} \text{ of SU(6)}
= ({\bf 2}, {\bf 1}, {\bf 1},1,1) ⊕({\bf 1},{\bf 2}, {\bf 1}, −1,1)
 ⊕ ({\bf 1},{\bf 1},{\bf 2},0, −2)
\text{ of SU(2) $\times$ SU(2) $\times$ SU(2) $\times$ U(1) $\times$ U(1)}.
$$
if we look at the SU(6) branch rule to SU(2) $\times$ SU(2) $\times$ SU(2),
we see that
$$
{\bf 6} \text{ of SU(6)}
= ({\bf 2}, {\bf 1}, {\bf 1}) ⊕({\bf 1},{\bf 2}, {\bf 1})
 ⊕ ({\bf 1},{\bf 1},{\bf 2})
\text{ of SU(2) $\times$ SU(2) $\times$ SU(2)}.
$$
There are indeed several SU(4) subgroups that we can find,
(1) by identifying the
$$
\Big(({\bf 2}, {\bf 1}, {\bf 1}) ⊕({\bf 1},{\bf 2}, {\bf 1})\Big)
 ⊕ ({\bf 1},{\bf 1},{\bf 2})
\text{ of SU(2) $\times$ SU(2) $\times$ SU(2) as }
({\bf 4}, {\bf 1})
 ⊕ ({\bf 1},{\bf 2})
\text{ of First SU(4) $\times$ SU(2)}
.
$$
(2) by identifying the
$$
({\bf 2}, {\bf 1}, {\bf 1}) ⊕\Big( ({\bf 1},{\bf 2}, {\bf 1})
 ⊕ ({\bf 1},{\bf 1},{\bf 2})\Big)
\text{ of SU(2) $\times$ SU(2) $\times$ SU(2) as }
({\bf 2}, {\bf 1})
 ⊕ ({\bf 1},{\bf 4})
\text{ of SU(2) $\times$ Second SU(4)'}
.
$$
So the SU(6) does contain the First SU(4) and the Second SU(4)' that you described, as long as we focus on the representation constraint on the ${\bf 4}$ of the First SU(4) and the Second SU(4)'.
So $$G= \text{ SU(6)}$$ so far.
However, it looks to me that you further impose extra conditions on the  ${\bf 6}$ of the First SU(4) and the Second SU(4)'. I am not 100% sure that the $G= \text{ SU(6)}$ works for the extra condition. It may overkill the solution so that there may be no solution at all.
