Every conformal vector field on $\mathbb{R}^n$ is homothetic? Consider $\mathbb{R}^n$ (for $n \ge 3$). Is it true that every conformal vector field on $\mathbb{R}^n$ is homothetic?
A vector field is homothetic if its flow is a homothety- a conformal map with a constant conformal factor. Equivalently, $V \in \Gamma(T\mathbb{R}^n)$ is homothetic if and only if
$$ \lambda \cdot g(X,Y)=g(\nabla_X V,Y)+g(X,\nabla_Y V),$$
where $g$ is the Euclidean metric, and $\lambda \in \mathbb{R}$.
I know the Killing fields are exactly those of the form $$ X(x)=Ax+b, \, \text{ where } \, A \, \text{ is a skew symmetric matrix and } \, b \in \mathbb{R}^n.  $$
Is there an explicit description of the conformal algebra of $\mathbb{R}^n$? (somehow I am having trouble finding a reference).
I hope for an elementary argument. (I thought to use Liouville's theorem but this seems an overkill, and I am not sure my use is correct).
 A: I am writing an answer using Moishe Cohen's comment:
Let $b \in \mathbb{R}^n$, and define $V:\mathbb{R}^n \to \mathbb{R}^n$ by
$$ V(x)=b|x|^2-2 \langle b,x \rangle x.$$
We shall show $V$ is a conformal vector field which is not a homothety. Since we are on $\mathbb{R}^n$, $\nabla V=dV$ is the standard derivative.
$$ dV_x(y)=2\langle x,y \rangle b-2 \langle b,y \rangle x-2\langle b,x \rangle y$$
If we write $S(y)=\langle b,y \rangle x$, then $S^T(y)=\langle x,y \rangle b$, so
$$ (dV_x)^T(y)=2\langle b,y \rangle x-2 \langle x,y \rangle b-2\langle b,x \rangle y.$$
This implies
$$ dV_x(y)+(dV_x)^T(y)=-4\langle b,x \rangle y,$$ i.e.
$$ dV_x+(dV_x)^T=-4\langle b,x \rangle \text{Id}.$$
This implies $V$ is conformal, since the conformal equation is $$ \nabla V+(\nabla V)^T=h\text{Id}_{TM},$$
where $h \in C^{\infty}(M)$. 
(By taking traces of both sides, one sees that $h=\frac{2}{n} \text{tr}(\nabla V)=\frac{2}{n} \text{div} V$).
So here, $\text{div} V=-2n\langle b,x \rangle$ is not constant, so $V$ is not homothetic.
