The paths of $UX_t$ are continuous and start at zero. Its increments are independent because they are deterministic functions of the independent increments of $X$. They are centered Gaussian, and orthogonality gives
$$
\operatorname{Cov}(U(X_t-X_s))
=U((t-s)I_d)U^T
=(t-s)I_d.
$$
Thus $UX$ is Brownian motion. This is the <orthogonal invariance of Brownian motion>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 201 / 4 / b / Solution

If $B$ is Brownian motion in $\mathbb R^d$ and $U$ is an orthogonal matrix, then $UB$ is Brownian motion. Its increments remain independent centered Gaussian vectors and have covariance $(t-s)UIU^T=(t-s)I$.


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 Incoming links: Orthogonal invariance of Brownian motion