Is there a unique "natural" action of $\mathsf{SL}_{n+1}$ on $\mathbb{R}^n$? Context
By acting naturally via $\mathsf{SL}_3$ on $\mathbb{RP}^2=\{[x:y:z]\}$ and by taking the induced action on the affine hyperplane $z=1$ (which we identify with $\mathbb{R}^2$), one can realize the corresponding Lie algebra $\mathfrak{sl}_3$ as a sub-algebra of the algebra $\mathfrak{X}(\mathbb{R}^2)$ of vector fields of $\mathbb{R}^2$: by carrying out this simple exercise, one ends up with 8 vector fields, 6 of which are linear, and two contain quadratic coefficients, namely
$$
X_A=-xy\partial_x-y^2\partial_y\, ,\quad X_B=-x^2\partial_x-xy\partial_y\, ,
$$
that correspond to the matrices
$$
A=\left(\begin{array}{ccc}0&0&0\\ 0&0&0\\0&1&0\end{array}\right)\,,\quad B=\left(\begin{array}{ccc}0&0&0\\ 0&0&0\\1&0&0\end{array}\right)\, ,
$$
respectively. Image below shows how to use the flow of $X\in\mathfrak{sl}_3$ to obtain a vector field on the plane $z=1$:

The 6 linear vector fields are of course complete, whereas the flows of $X_A$ and $X_B$ display a singularity at $t=-\tfrac{1}{y}$ and $t=-\tfrac{1}{x}$, respectively. This remark motivated the first question:
For dimensions reasons (as observed in the comments), for every $n\ge 1$ there is no embedding
$$
\mathfrak{sl}_{n+1}\hookrightarrow\mathfrak{X}(\mathbb{R}^n)\, ,
$$
whose image is made of linear (allowing a constant) vector fields.
However for any $n\geq 1$ it is easy to find an embedding
$$
i_{\textrm{nat}}:\mathfrak{sl}_{n+1}\hookrightarrow\mathfrak{X}(\mathbb{R}^n)\, ,
$$
which we may call natural, because it is induced by the natural action of $\mathsf{SL}_{n+1}$ on $\mathbb{RP}^n$ by the same construction as above, one may wonder the following (initially "Question 2"):
Question
Given an arbitrary embedding
$$
i :\mathfrak{sl}_{n+1}\hookrightarrow\mathfrak{X}(\mathbb{R}^n)\, ,
$$
it is true that $i$ must necessarily be equivalent to $i_{\textrm{nat}}$? If not, which conditions must $i$ satisfy, to be such?
In other words, one has a collection $X_1,X_2,\ldots, X_{d(n)}$, where $d(n)=(n+1)^2-1=\dim \mathsf{SL}_{n+1}$ of vector fields on $\mathbb{R}^n$, that commute according to the commutation relations of $\mathsf{SL}_{n+1}$ and the question can be recast as follows: is there a diffeomorphism of $\mathbb{R}^n$ that transforms the aforementioned vector fields into the images via $i_{\textrm{nat}}$ of a suitable set of generators of $\mathfrak{sl}_{n+1}$? What kind of obstruction (if any) one should expect?
 A: 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

*

*$\dim L=1$ and, up to local diffeomorphism, $L$ is spanned by $\frac{\partial}{\partial x}$,


*$\dim L=2$ and, up to local diffeomorphism, $L$ is spanned by $\frac{\partial}{\partial x},\ x\,\frac{\partial}{\partial x}$,


*$\dim L=3$ 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


*$\dim L=\infty$, 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.  A famous example is the split form of type $\mathfrak{c}_2$ (usually denoted as $\mathfrak{sp}_4(\mathbb{R})$), a Lie algebra of dimension $10$ that has two non-conjugate families of subalgebras of minimal codimension $3$, leading to two inequivalent ways that it can appear as a Lie algebra of vector fields on a $3$-manifold.  One is as contact vector fields on $\mathbb{RP}^3$, and the other is as the conformal vector fields on $\mathcal{N}^{2,1}$, the manifold of null subspaces of dimension $1$ in Minkowski $5$-space, $\mathbb{M}^{4,1}$.  This is the basis of the classical Klein correspondence.
For more exotic examples, 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 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 $\mathbb{R}^{15}$.
