Explaining a simple observation on Terry Tao's blog about the Wiener process Quoting Terry Tao's blog:

A simple but fundamental observation is that $n$-dimensional Brownian
  motion is rotation-invariant: more precisely, if $(X_t)_{t \in
 [0,+\infty)}$ is an $n$-dimensional Wiener process with initial
  position $0$, and $U \in O(n)$ is any orthogonal transformation on
  ${\bf R}^n$, then $(UX_t)_{t \in [0,+\infty)}$ is another Wiener
  process with initial position $0$, and thus has the same distribution:
$(UX_t)_{t \in [0,+\infty)} \equiv (X_t)_{t \in [0,+\infty)}$.
This is ultimately because the $n$-dimensional normal distributions $N(0,\sigma^2 I)_{{\bf R}^n}$ are manifestly rotation-invariant (see Exercise 10 of Notes 1).

I do not have the solution to Exercise 10 and I do not see directly how it's related. Could someone clarify in more detail how this explains that Brownian motion is rotation-invariant? (To show that all the conditions don't change).
 A: Let $W_t$ be an $n$-dimensional Brownian motion. Then, we of course know that $t \mapsto W_t(\omega)$ is continuous everywhere for every $\omega$ in a full measure set. $W_0 = 0$ a.s. and $[W^i,W^j]_t = \delta_{ij}t$ where $W^i$ is the $i^{\text{th}}$ coordinate of $W$ and $\delta_{ij}$ is the Kronecker delta. Finally, $(W^i_t)_t$ is a martingale for every $i=1,\ldots,n$.
Define $X_t(\omega) = RW_t(\omega)$ for every $t$ and $\omega$ where $R$ is a rotation matrix. For every $\omega$ such that $t \mapsto W_t(\omega)$ is continuous, $t \mapsto RW_t(\omega)$ is continuous as well. $X_0 = 0$ a.s. also.
Let $X^j$ denote the $j^{\text{th}}$ coordinate of the process $X$. Then, $X^j_t = R^jW_t$ where $R_j$ is the $j^{\text{th}}$ row of $R$. For $s \leq t$,
$$E[X^j_t \mid \mathcal{F}_s] = E[R^jW_t \mid \mathcal{F}_s] = R^jE[W_t \mid \mathcal{F}_s] = R^jW_s =: X^j_s$$
So $X$ is a martingale itself.
Let $R^i_k$ denote the row $i$, column $k$ element of $R$.
$$[X^i,X^j]_t = [R^iW,R^jW]_t = [\sum_{k=1}^nR^i_kW^k,\sum_{k=1}^nR^j_kW^k]_t = t\sum_{k=1}^nR^i_kR^j_k$$
Since $R$ is a rotation matrix, $R^{\top}R = I$. Hence $\sum_{k=1}^nR^i_kR^j_k =  \delta_{ij}$.
Then, by Levy's characterization we know $X$ must be BM.
