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 $n^1,n^2,n^3$) $$ {\bf 3} \text{ as } V^j := \Big( u^{*T} (n^1 \sigma^1) u, \quad u^{*T} (n^2 \sigma^2) u, \quad u^{*T} (n^3 \sigma^3) u\Big) $$
(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 $V^j := \big(u^{*T} (n^j \sigma^j) u\big)$ 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 $V^j := \big(u^{*T} (n^j \lambda_j) u\big)$ with $j=1,2,3, \dots, 8$invariant?
I believe the question that you are asking may be rephrased as:
Given a vector $v$ in the irreducible real 8-dimensional representation of $\mathrm{SU}(3),$ determine the subgroup $\mathrm{G} < \mathrm{SU}(3)$ stabilising $v.$
The irreducible real 8-dimensional representation of $\mathrm{SU}(3)$ is the adjoint representation $\mathfrak{su}(3)$ (you will need to multiply the Gell-Mann matrices by $i$). By Cartan's theorem on maximal tori, every element $v \in \mathfrak{su}(3)$ is conjugate to an element of the form $$u(\lambda_1, \lambda_2) = \begin{pmatrix} i \lambda_1 & 0 & 0 \\ 0 & i \lambda_2 & 0 \\ 0 & 0 & -i (\lambda_1 + \lambda_2) \end{pmatrix}, \:\:\: 0 \leq \lambda_1 \leq \lambda_2.$$
To determine $\lambda_1$ and $\lambda_2$ from $v,$ compute $$|v|^2 = 2 \lambda_1^2 + 2 \lambda_1 \lambda_2 + 2 \lambda_2^2$$ and $$* (v^{\flat} \wedge v^{\flat} \wedge v^{\flat}) = -6 \lambda_1 \lambda_2 (\lambda_1 + \lambda_2)$$ and solve for $\lambda_1$ and $\lambda_2.$
The $\mathrm{SU}(3)$-stabiliser $\mathrm{G}$ of a nonzero $v$ depends only on the latter quantity. If $\lambda_1 \lambda_2 (\lambda_1 + \lambda_2) \neq 0,$ then $\mathrm{G} \cong \mathrm{T}^2$ is a maximal torus in $\mathrm{SU}(3).$ If $\lambda_1 \lambda_2 (\lambda_1 + \lambda_2) = 0,$ then $\mathrm{G}$ is conjugate to the subgroup $\mathrm{S}(\mathrm{U}(2) \mathrm{U}(1) ) < \mathrm{SU}(3)$ of matrices of the form $$\begin{pmatrix} A & 0 \\ 0 & (\mathrm{det} A)^{-1} \end{pmatrix}, \:\:\: A \in \mathrm{U}(2).$$
The analogue of the fibration structure in the toy examples is the cohomogeneity-one action of $\mathrm{SU}(3)$ on the unit sphere $S^7 \subset \mathfrak{su}(3).$