This probably should only require elementary differential geometry but somehow I can't seem to prove it.
How can I show the vector field $\displaystyle \sum_{i=1}^n x_i \frac{\partial}{\partial x_i}$ on $\mathbb{R}^n$ is invariant under the action of $\mathrm{SO}(n)$?
On
To complement Mike's answer in terms of the flow, one can show this directly via the vector field, call it $X := \sum_i x_i \frac{\partial }{\partial x_i}$, by considering a linear map $A \in \text{GL}(n, \mathbb{R})$ and noting
\begin{align*} (A_*X)_{Ax} = & A_* (X_{x}) = \sum_i x_i A_* \frac{\partial}{\partial x_i} \\ = & \sum_{i,j} x_i \frac{\partial (Ax)_j}{\partial x_i} \frac{\partial}{\partial x_j} \\ = & \sum_{i,j} x_i A_{ji} \frac{\partial}{\partial x_j} \\ = & \sum_{j} (Ax)_j \frac{\partial}{\partial x_j} \\ = & X_{Ax}, \end{align*}
Showing $A_* X = X$. Another way (practically the same as the flow observation) is to note that a curve $\gamma_x : \mathbb{R} \to \mathbb{R}^n$ representing the vector $X_x$ (meaning $\gamma_x'(0) = X_x$) is $\gamma_x (t) = (1+t)x$, so that $$(A \circ \gamma_x)(t) = A \gamma_x (t) = (1+t)Ax = \gamma_{Ax}(t),$$ and hence $$(A_*X)_{Ax} = (A \circ \gamma_x)'(0) = \gamma_{Ax}'(0) = X_{Ax}.$$
This is called the "Euler vector field" and it is actually invariant under all of $\mathrm{GL}(n,\mathbb{R})$. A nice way to see this is to look at its flow which is the scaling action $$ \phi_t(x) = e^t x.$$ It is quite clear that this flow is invariant, because linear maps commute with scalar multiplication, and this is equivalent to invariance of the generating vector field.