Invariant $SU(3)$ subgroup for ${bf 8}$ in ${bf 3}^* otimes {bf 3} ={bf 1} oplus {bf 8}$

This question concerns finding an invariant subgroup of a total group $G$.

Warm-Up Toy Example (which I already solved):

Let us take $G=SU(2)$ as a special unitary group. Let us take $$u text{ as }
{bf 2} text{ of } G=SU(2)$$ has the fundamental representation ${bf 2}$ (two complex components) of $G=SU(2)$. There is also the complex conjugate representation $u^*$, but in this case, ${bf 2}^*={bf 2}$ still.

Now we can combine two fundamental representations to form
$$
{bf 2} otimes {bf 2} = {bf 2}^*
otimes {bf 2} ={bf 1} oplus {bf 3}.
$$
Where
$$
{bf 2} otimes {bf 2} = {bf 2}^*
otimes {bf 2} =u^{*T} (n^j sigma^j) u.
$$
Again $*$ is complex conjugation, and $T$ is the transpose.
with $sigma^j$ of $j=0,1,2,3$ and https://en.wikipedia.org/wiki/Pauli_matrices
$$
sigma^0=begin{pmatrix}
1&0\
0&1
end{pmatrix},
sigma^1=begin{pmatrix}
0&1\
1&0
end{pmatrix},
sigma^2=begin{pmatrix}
0&-i\
i&0
end{pmatrix},
sigma^3=begin{pmatrix}
1 &0\
0 &-1
end{pmatrix}.
$$
So for a fix $j$, $u^{*T} (n^j sigma^j) u$ outputs a scalar number.

Here we can define a scalar (for some coefficient $n^0$)
$$
{bf 1} text{ as } V^0 := u^{*T} (n^0 sigma^0) u
$$
and a 3-vector (for some coefficient $n^1,n^2,n^3$)
$$
{bf 3} text{ as } V^j := Big( u^{*T} (n^1 sigma^1) u, quad u^{*T} (n^2 sigma^2) u, quad u^{*T} (n^3 sigma^3) uBig)
$$

(warm-up 1) If we choose to look at ${bf 1}$ as $V^0$ with a fixed $n^0$ coefficient, we can ask:

What are the subgroup of $SU(2)$ which acts on $u mapsto u’=exp(i theta^a sigma^a) u$
makes the ${bf 1}$ as $V^0:= u^{*T} (n^0 sigma^0) u$ invariant?

Answer: The full $SU(2)$ makes $V^0$ invariant. Since under $SU(2)$ transformation:
$$V^0:= u^{*T} (n^0 sigma^0) u mapsto
{V^0}’=
{u’}^{*T} (n^0 ) u’ = u^{*T} exp(-i theta^a sigma^a)
(n^0 sigma^0) exp(i theta^a sigma^a) u
=
u^{*T}
(n^0 ) u = V^0$$
for arbitrary $theta^j$, thus the full $SU(2)$.
In short, there is a fibration structure:
$$
text{stablizer $hookrightarrow$ total $G$ $to$ orbit}.
$$
From the $SU(2)$ view:
$$
SU(2)hookrightarrow SU(2) to pt
$$

(warm-up 2) If we choose to look at ${bf 3}$ as $V^j$ with a fixed coefficient $n^1,n^2,n^3$ coefficient, we can ask:

What are the subgroup of $SU(2)$ which acts on $u mapsto u’=exp(i theta^a sigma^a) u$
makes the ${bf 3}$ as $V^j := big(u^{*T} (n^j sigma^j) ubig)$ with $j=1,2,3$invariant?

Answer: Only the $U(1)$ subgroup of $SU(2)$ makes $V^j$ invariant. Since under $SU(2)$ transformation:
$$V^j:= u^{*T} (n^j sigma^j) u mapsto
{V^j}’=
{u’}^{*T} (n^j sigma^j) u’ = u^{*T} exp(-i theta^a sigma^a)
(n^j sigma^j) exp(i theta^a sigma^a) u.
$$
The $V^j$ is a 3-vector on an $S^2$ sphere. while the $SU(2)$ acts on $u$ becomes effectively as
$SO(3)$ acts on $V^j$ on the $S^2$ sphere. With a fixed coefficient $n^1,n^2,n^3$ coefficient,
only when the $SO(3)$ subgroup that makes the $V^j$ 3-vector on an $S^2$ sphere invariant would be the desired subgroup, which is the $U(1)$ subgroup.

In short, there is a fibration structure:
$$
text{stablizer $hookrightarrow$ total $G$ $to$ orbit}.
$$
From the $SO(3)$ view:
$$
U(1)hookrightarrow SO(3) to S^2
$$
From the $SU(2)$ view:
$$
Spin(2)hookrightarrow SU(2) to S^2
$$
such that $Spin(2)/mathbb{Z}_2=U(1)$.

Serious Puzzle:

Let us take $G=SU(3)$ as a special unitary group. Let us take $$u text{ as }
{bf 3} text{ of } G=SU(3)$$ has the fundamental representation ${bf 3}$ (three complex components) of $G=SU(3)$. There is also the complex conjugate representation $u^*$, but in this case, ${bf 3}^*$ which is distinct from ${bf 3}$.

Now we can combine two fundamental representations to form
$$
{bf 3}^* otimes {bf 3} ={bf 1} oplus {bf 8}.
$$
We can take $$
{bf 3}^*
otimes {bf 3} =u^{*T} (n^j lambda_j) u.
$$
with the https://en.wikipedia.org/wiki/Gell-Mann_matrices:

$lambda_0 = begin{pmatrix} 1 & 0 & 0 \ 0 &1 & 0 \ 0 & 0 & 1 end{pmatrix}$,
$lambda_1 = begin{pmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 0 end{pmatrix}$,
$lambda_2 = begin{pmatrix} 0 & -i & 0 \ i & 0 & 0 \ 0 & 0 & 0 end{pmatrix}$,
$lambda_3 = begin{pmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 0 end{pmatrix}$,
$lambda_4 = begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 1 & 0 & 0 end{pmatrix}$,
$lambda_5 = begin{pmatrix} 0 & 0 & -i \ 0 & 0 & 0 \ i & 0 & 0 end{pmatrix}$,
$lambda_6 = begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 end{pmatrix}$,
$lambda_7 = begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & -i \ 0 & i & 0 end{pmatrix}$,
$lambda_8 = frac{1}{sqrt{3}} begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -2 end{pmatrix}$
with $j=1,2,3, dots, 8$

(warm-up 3) If we choose to look at ${bf 1}$ as $V^0$ with a fixed $n^0$ coefficient, we can ask:

What are the subgroup of $SU(3)$ which acts on $u mapsto u’=exp(i theta^a lambda_a) u$
makes the ${bf 1}$ as $V^0:= u^{*T} (n^0 lambda_0) u$ invariant?

The answer shall be the full $SU(3)$.

(FINALY NOW Puzzle)If we choose to look at ${bf 8}$ as $V^j$ with a fixed coefficient $n^1,n^2,n^3,dots, n^8$ coefficient, we can ask:

What are the subgroup of $SU(3)$ which acts on $u mapsto u’=exp(i theta^a lambda_a) u$
makes the ${bf 8}$ as $V^j := big(u^{*T} (n^j lambda_j) ubig)$ with $j=1,2,3, dots, 8$invariant?