There are two aspects to this question, the global question and the local question. Also, the case $n=1$ is different from $n>1$. Basically, the answer is 'essentially yes, but with some caveats'.
Here's a sample of the kind of results one might consider as an answer in the case $n=1$:
First, an old theorem of Sophus Lie:
*Theorem:* (Lie) Let $L\subset\mathfrak{X}(\mathbb{R}^1)$ be a Lie algebra of vector fields that is *locally transitive*, i.e., for every $p\in\mathbb{R}$, there is a vector field $X\in L$ such that $X(p)\not=0$. Then either
1. $\dim L=1$ and, up to local diffeomorphism, $L$ is spanned by $\frac{\partial}{\partial x}$,
2. $\dim L=2$ and, up to local diffeomorphism, $L$ is spanned by $\frac{\partial}{\partial x},\ x\,\frac{\partial}{\partial x}$,
3. $\dim L=3$ and, up to local diffeomorphism, $L$ is spanned by $\frac{\partial}{\partial x},\ x\,\frac{\partial}{\partial x},\ x^2\,\frac{\partial}{\partial x}$, or
4. $\dim L=\infty$, and for every $k$ and every $p\in\mathbb{R}$, there is a vector field $X_k\in L$ such that $X_k$ vanishes to order exactly $k$ at $p$.
Then there is a global result that is not too hard to prove:
*Theorem:* Suppose that $L\subset\mathfrak{X}(\mathbb{R})$ is a locally transitive Lie algebra of dimension $3$. Then there is a diffeomorphism $u:\mathbb{R}\to u(\mathbb{R})\subseteq\mathbb{R}$ that carries $L$ into the span of the restrictions to $u(\mathbb{R})$ of the vector fields $\frac{\partial}{\partial x}$, $\cos x\,\frac{\partial}{\partial x}$, and $\sin x\,\frac{\partial}{\partial x}$.
Note that all the vector fields in $L$ are complete if and only if $u(\mathbb{R})=\mathbb{R}$. In particular, if $L\subset\mathfrak{X}(\mathbb{R})$ is the span of $\frac{\partial}{\partial x}$, $\cos x\,\frac{\partial}{\partial x}$, and $\sin x\,\frac{\partial}{\partial x}$, then there is an $L$-preserving diffeomorphism $\phi:(a,b)\to(c,d)$ between two bounded intervals $(a,b),(c,d)\subset\mathbb{R}$ if and only if either $b-a=d-c=2\pi k$ or else $2\pi(k{-}1) < b-a, d-c <2\pi k$ for some integer $k\ge1$.
Thus, this gives a complete answer in the case $n=1$ and the locally transitive case. The need for the hypothesis of local transitivity is shown by this example: Let $\phi:\mathbb{R}\to [0,1)$ be a smooth function such that $\phi(x)>0$ when $|x|<1$ and $\phi(x)=0$ when $|x|\ge 1$. Let $u:(-1,1)\to \mathbb{R}$ satisfy $u'(x) = 1/\phi(x)$ when $|x|<1$. Then $u:(-1,1)\to\mathbb{R}$ is a diffeomorphism. Let $L\subset\mathfrak{X}\bigl((-1,1)\bigr)$ be the Lie algebra spanned by the vector fields
$$
\phi(x)\,\frac{\partial}{\partial x},\quad
\cos\bigl(u(x)\bigr)\,\phi(x)\,\frac{\partial}{\partial x},\quad
\sin\bigl(u(x)\bigr)\,\phi(x)\,\frac{\partial}{\partial x}.
$$
These vector fields extend smoothly to the entire real line as zero vector fields where $|x|\ge1$, but one could also extend them smoothly to be periodic of period $2$. In either case, they vanish to infinite order at the odd integers. Clearly, there are infinitely many globally inequivalent ways to embed $\mathfrak{sl}_2$ as a subalgebra of $\mathfrak{X}(\mathbb{R})$ if one does not assume local transitivity. Classification is pretty hopeless.
Now assume that $n>1$. Here, the situation is somewhat more straightforward. The first thing to notice is the classical result that $L=\mathfrak{sl}_{n+1}=\mathfrak{sl}_{n+1}(\mathbb{R})$ has no proper subalgebras of codimension less than $n$. (This was known to Lie, at least in the complex case, i.e., for $\mathfrak{sl}_{n+1}(\mathbb{C})$, and the result for the real case follows immediately from this by complexification.) In fact, there is a complete descriptions of the set $S_n$ of subalgebras of $L$ that have codimension $n$. This is an $n$-manifold with two components
$$
S_n = \mathrm{SL}(n{+}1,\mathbb{R})/P_1 \cup
\mathrm{SL}(n{+}1,\mathbb{R})/P_2\simeq \mathbb{RP}^n\cup \mathbb{RP}^n\,
$$
where $P_1$ is the maximal parabolic consisting of the elements of $\mathrm{SL}(n{+}1,\mathbb{R})$ that preserve a given $1$-dimensional subspace of $\mathbb{R}^{n+1}$ and $P_2$ is the maximal parabolic consisting of the elements of $\mathrm{SL}(n{+}1,\mathbb{R})$ that preserve a given $n$-dimensional subspace of $\mathbb{R}^{n+1}$. (These two subgroups are not conjugate in $\mathrm{SL}(n{+}1,\mathbb{R})$, but there is an outer automorphism ($a\mapsto (a^{-1})^\mathsf{T}$) that carries one to the other. (Of course, when $n=1$, the space $S_1=\mathbb{RP}^1\simeq S^1$ has only one component.)
Now, suppose that $L$ is embedded as a subalgebra of $\mathfrak{X}(M^n)$, where $M^n$ is a smooth manifold of dimension $n$. Then, for any $p\in M$, the subalgebra $L^0_p\subset L$ consisting of the vector fields that vanish at $p$ has codimension at most $n$, so either $L^0_p = L$, in which case, all the vector fields in $L$ vanish at $p$, or else the codimension of $L^0_p$ in $L$ is exactly $n$. Let's remove the closed set of points where all the vector fields in $L$ vanish (which will not affect the completeness of any of the vector fields in $L$) and assume that $L^0_p$ always has codimension $n$.
Thus, we have a canonical map $\phi:M\to S_n$ given by $\phi(p) = L^0_p$.
*Theorem:* If $L\subset\mathfrak{X}(M^n)$ is a locally transitive subalgebra isomorphic to $\mathfrak{sl}_{n+1}$, then $\phi:M\to S$ is a smooth immersion (in particular, $\phi$ is a local diffeomorphism). Moreover, $\phi_*$ carries $L$ to the 'natural' copy of $L$ in $\mathfrak{X}(S_n)$ induced by the action of $\mathrm{SL}(n{+}1,\mathbb{R})$ on $S_n$.
The proof of this theorem is not hard, but it consists of a number of steps, whose details would take up a lot of space. The main points are these: First, let $L^i_p\subset L$ denote the subset of vector fields that vanish to order $i{+}1$ at $p$, then one first shows that $L^i_p = 0$ for $i$ suffciently large (*a priori* depending on $p$). Then, using the fact that $L^0_p$ contains a simple subalgebra isomorphic to $\mathfrak{sl}_n$, one uses its representation theory to show that this simple subalgebra cannot be contained in $L^1_p$ (which is a solvable ideal of $L^0_p$), and hence $L^0_p/L^1_p$ is an algebra containing $\mathfrak{sl}_n$. Using the nondegeneracy of the Killing form, one concludes that $L^1_p$ is dual, as a $\mathfrak{sl}_n$-representation, to $L/L^0_p\simeq\mathbb{R}^n$ and hence has dimension $n$. Thus, $L^0_p/L^1_p$ has dimension $n^2$ and must be isomorphic to $\mathfrak{gl}_n$. From this, it follows easily that $L^2_p = (0)$. Now, using these facts, it is easy to explicitly compute the differential of $\phi$ at $p$ and show that $\phi'(p):T_pM\to T_{\phi(p)}S_n$ is an isomorphism. Once one has the fact that $\phi$ is a local diffeomorphism, the final statement follows relatively easily.
Note that because $S_n$ has two components there are essentially two inequivalent global realizations of $L=\mathfrak{sl}_{n+1}$ as vector fields on $\mathbb{RP}^n$. By this, I mean that there are two Lie algebra homomorphisms $\psi_i:L\to\mathfrak{X}(\mathbb{RP}^n)$ such that there is no diffeomorphism $u:\mathbb{RP}^n\to\mathbb{RP}^n$ such that $\psi_2(X) = u_*\bigl(\psi_1(X)\bigr)$ for all $X\in L$. What is true instead is that there is an outer automorphism $\tau:L\to L$ such that $\psi_2\bigl(\tau(X)\bigr) = u_*\bigl(\psi_1(X)\bigr)$. This is, perhaps, a subtle point, but it shows that there really are two essentially different ways that $L$ can be realized as vector fields in dimension $n$. It is not clear which one should be called 'natural'.
Also, note that, if $L\subset\mathfrak{X}(M)$ consists entirely of complete vector fields, then $\phi:M\to S_n$ must be a covering space, in particular, if $M$ is connected, then it must be either $S^n$ or $\mathbb{RP}^n$. (Note that this is another place where $n>1$ differs from the case $n=1$.)
*Remark:* This non-uniqueness can be even more dramatic. For example, the split form of the exceptional Lie algebra $\mathfrak{g}_2$ (of dimension $14$) has two non-conjugate subalgebras of the minimal codimension $5$. Since $\mathfrak{g}_2$ has no outer automorphisms, they are not even equivalent up to isomorphisms of the algebra. Thus, as both Cartan and Engel realized in 1893, $\mathfrak{g}_2$ can appears as two essentially different subalgebras of the vector fields in $\mathbb{R}^5$.
This phenomenon also happens for the split form of the exceptional Lie algebra $\mathfrak{f}_4$, and, correspondingly, it can be realized as two essentially different subalgebras of the vector fields in $\mathbb{R}^{15}$.