If we are given a Lie algebra of vector fields $\{X_i\}_{i = 1} ^N$ on a manifold $M ^n$, is it possible to determine in local coordinates the $X_i$'s if we know the value of $m := \dim \text{span} \{X_i\}_{i = 1} ^N$?
As a first example, consider $N= 2 = m$, and $[X_1, X_2] = 0$. Then by Frobenius we can find coordinates s.t. $X_i = \partial_i$, so in this case it is indeed possible to determine $X_1$ and $X_2$. If we had instead assumed that $[X_1, X_2] = X_1$, Frobenius allows us to represent $X_1 = \partial_1$ and $X_2 = a \partial_1 + b \partial_2$. The structure equations then show that $a = x ^1 + c(x ^2)$ and $b = b(x ^2)$. Changing coordinates via \begin{align*} x ^1 &= y ^1 + f(y ^2), \\ x ^2 &= g(y ^2) \end{align*} transforms $X_1$ and $X_2$ into \begin{align*} \tilde{X}_1 &= \partial_1, \\ \tilde{X}_2 &= (y ^1 + f(y ^2) + c(g(y ^2)) + f'(y ^2)b(g(y ^2)))\partial_1 + b(g(y ^2))g'(y ^2) \partial_2. \end{align*} Since $X_1$ and $X_2$ are everywhere independent, $b$ is never zero, so on any small enough interval we can solve $g'(y ^2) = 1/b(g(y ^2))$ and then find $f$ solving the linear ODE $$ b(g(y ^2))f'(y ^2) + f(y ^2) + c(g(y ^2)) = 0, $$ implying $\tilde{X}_2 = y ^1 \partial_1 + \partial_2$. In other words, for $N = 2 = m$, we can indeed determine the $X_i$'s from just the structure equations.
However, consider the case $N = 3$, $m = 2$, and \begin{align*} [X_1, X_2] &= X_3, \\ [X_2, X_3] &= X_1, \\ [X_3, X_1] &= X_2. \end{align*} By the Frobenius theorem, we can find a coordinate system $(x ^i)$ s.t. $$ X_1 = \partial_1, \quad X_2 = a \partial_1 + b \partial_2, \quad X_3 = c \partial_1 + d \partial_2. $$ The structure equations $[X_1, X_2] = X_3$ and $[X_1, X_3] = -X_2$ force \begin{align*} \partial_1 a &= c, \quad -\partial_1 c = a \Rightarrow \partial_1 ^2 a + a = 0, \\ \partial_1 b &= d, \quad - \partial_1 d = b \Rightarrow \partial_1 ^2 b + b = 0, \end{align*} from which we deduce that $a = a_1 \sin(x ^1) + a_2 \cos(x ^1)$ and $b = b_1 \sin(x ^1) + b_2 \cos(x ^1)$. Plugging this into $[X_2, X_3] = X_1$ yields \begin{align*} b_2 \partial_2a_1-b_1 \partial_2a_2-a_1^2-a_2^2&=1, \\ b_2 (\partial_2b_1-a_2)-b_1 (a_1+\partial_2b_2) &= 0. \end{align*} One can play around with different guesses for $a_1$ and $b_1$ to see that there are a wealth of solutions to these equations. Since there are no derivatives $\partial_{x ^i}$ with $i > 2$, we further have great freedom in choosing the ''constants of integration'', which only results in even more families of solutions. It is not obvious how to choose new coordinates which ensure all of these families are equivalent like in the previous example.
Thus, I am left with the following
Question
What additional information on the $X_i$'s would be required to single out a canonical solution in general or ensure all solutions are equivalent (e.g. via a change of variables)?
In short, no: Already for $N = 2$, $m = 3$ we can compute that the real spans of $$\{\partial_x, x\,\partial_x + y \partial_y, (x^2 - y^2) \partial_x + 2 x y \partial_y\}, \qquad \{\partial_x, x \partial_x, x^2 \partial_x\}$$ are both isomorphic as Lie algebras to $\mathfrak{sl}_2(\Bbb R)$, but there is no local diffeomorphism of $\Bbb R^2$ that maps one Lie algebra to the other: The latter is imprimitive, i.e., it admits a $1$-dimensional foliation invariant under the action of the algebra, but the former does not (i.e., it is primitive).
In fact there are two more imprimitive real Lie algebras of vector fields in the plane inequivalent under local diffeomorphism to the one above, namely the real spans of $$\{\partial_x + \partial_y, x \partial_x + y \partial_y, x^2 \partial_x + y^2 \partial_y\} \qquad \textrm{and} \qquad \{\partial_x, 2 x \partial_x + y \partial_y, x^2 \partial_x + x y \partial_y\} .$$
The classification of Lie algebras of real vector fields in the plane is due to González-López–Kamran–Olver, wherein the above four Lie algebras appear as entries (2), (11), (17), (18) in Table 1 (see the citation below, which links to a pdf of the article). The analogous classification in the complex setting dates back to Lie himself.
Remark In the particular example for $N = 2, m = 3$ that you mention, for the Lie algebra $\mathfrak{so}(3)$ (with generators $X_1, X_2, X_3$ satisfying $[X_2, X_3] = X_1$ and its cyclic permutations), there is only one Lie algebra of vector fields up to local diffeomorphism (i.e., "canonical coordinates"), namely, we can find local coordinates $(x, y)$ such that \begin{align*} X_1 &= y \partial_x - x \partial_y \\ X_2 &= \frac{1}{2}(x^2 - y^2 + 1) \partial_x + x y \partial_y \\ X_3 &= x y \partial_x + \frac{1}{2}(y^2 - x^2 + 1) \partial_y . \end{align*} This algebra of vector fields appears as entry (3) of the cited table (but they have been rescaled here to satisfy the above particular relations in the question statement).
As far as I know there isn't a satisfying general answer for what additional information is needed to pin down a particular local diffeomorphism type. The indicated classification shows that for most of the isomorphism types of Lie algebras realizable using vector fields in the plane there is only a single local diffeomorphism type, so in those cases no more information is needed. But that statement is particular to $N = 2$ (and relies on a 30-page classification), and less is known in higher dimensions. Probably the best one can attempt to do is handle each $N$ and isomorphism type separately, but in many cases the problem may be a difficult one.