I have an orthonormal basis $\{ b_{\{1, 2\}}, b_{\{1, 3\}}, b_{\{2, 3\}}\}$ of the Hermitian space $\mathbb{C}^{3}$, with Hermitian product denoted by $(\cdot, \cdot)$, and the induced norm by $|\cdot|$. The basis vectors are indexed by the size 2 subsets of $\{1, 2, 3\}$. Since these indices are sets, $b_{\{p, q\}} = b_{\{q, p\}}$.
We will use the tensor product notation to define linear operators on $\mathbb{C}^{3}$, as follows: if $u, v$ are vectors in $\mathbb{C}^{3}$, then the operator $u \otimes v$ is defined to send $x \in \mathbb{C}^{3}$ to $(v, x) u$.
Suppose we are given the one-parameter groups $$ g^{t}_{(1 \; 2)} = {1 \over 2} \left( \left(b_{\{3, 1\}} + b_{\{3, 2\}}\right)^{\otimes 2} + e^{i 2 \pi t} \left(b_{\{3, 1\}} - b_{\{3, 2\}}\right)^{\otimes 2} \right) + b_{\{1, 2\}}^{\otimes 2}, $$ $$ g^{t}_{(1 \; 3)} = {1 \over 2} \left( \left(b_{\{2, 1\}} + b_{\{2, 3\}}\right)^{\otimes 2} + e^{i 2 \pi t} \left(b_{\{2, 1\}} - b_{\{2, 3\}}\right)^{\otimes 2} \right) + b_{\{1, 3\}}^{\otimes 2}, $$ $$ g^{t}_{(2 \; 3)} = {1 \over 2} \left( \left(b_{\{1, 2\}} + b_{\{1, 3\}}\right)^{\otimes 2} + e^{i 2 \pi t} \left(b_{\{1, 2\}} - b_{\{1, 3\}}\right)^{\otimes 2} \right) + b_{\{2, 3\}}^{\otimes 2}, $$ acting on $\mathbb{C}^{3}$. (These groups are indexed by transpositions of pairs of elements in $\{1, 2, 3\}$.)
It can be shown by direct computation in the orthonormal basis $$ c_1 = b_{\{1, 2\}}, \; c_2 = {1 \over \sqrt{2}}\left(b_{\{3, 1\}} + b_{\{3, 2\}}\right), \; c_3 = {1 \over \sqrt{2}}\left(b_{\{3, 1\}} - b_{\{3, 2\}}\right), $$ the group $g^{t}_{(1 \; 2)}$ fixes $c_{1}, c_{2}$ and preserves the equality $|\left(x, c_{3}\right)| = $const. Namely, $$ g^{t}_{(1 \; 2)} c_3 = e^{i 2 \pi t} c_3. $$ It follows that $$ \left|\left(g^{t}_{(1 \; 2)} x, c_{3} \right) \right| = \left|\left(x, c_{3} \right) \right|. $$
Analogous statements and computations hold for the other two groups, implying that all three groups are subgroups of the unitary group $\mbox{U}\left(3, \mathbb{C}\right)$.
The following setting, and some of the notation, is that of the Nagano-Sussmann Theorem. Define the orbit of a point $x \in \mathbb{C}^{3}$ as the set $$ {\cal O}_{x} = \{ g^{t_{k}}_{\alpha_{k}} g^{t_{k-1}}_{\alpha_{k-1}} \ldots g^{t_{1}}_{\alpha_{1}} x \; : \; t_{j} \in \mathbb{R}, \alpha_{j} \in \{(1 \; 2), (1 \; 3), (2 \; 3)\}, k \in \mathbb{N}, j = 1, \ldots, k \}. $$ Let $\mathfrak{g}$ be the Lie algebra generated by the tangent vectors $$ \left.{d \over dt}\right|_{t=0} g^{t}_{\alpha}, $$ and let $\mathfrak{g}_{x}$ be the evaluation of this Lie algebra at $x$. Let $y$ be another point in ${\cal O}_{x}$. My question is: Does there necessarily exist an element in $\mathfrak{g}$ with the corresponding one-parameter group $g_{y}^{t}$, $$ y = g_{y}^{t} \, x \mbox{ for some } t>0? $$
It would seem that ${\cal O}_{x}$ is the 3-D sphere, but the evaluation of the Lie algebra generated by the $g^{t}$ is not of the same dimension at every $x \in \mathbb{C}^3$. Thoughts? Criticisms?