Yes.
If $v$'s are orthonormal then $\left[\matrix{v_1^\top \\ v_2^\top}\right]\left[\matrix{v_1 & v_2}\right]=\left[\matrix{1 & 0 \\ 0 & 1 }\right]$ and if the $u$'s are defined for some matrix $(a,b,c,d)$ $\left[\matrix{a & b \\ c & d}\right]\left[\matrix{v_1 & v_2}\right]=\left[\matrix{u_1 & u_2}\right]$ and $u$'s are orthonormal $\left[\matrix{u_1^\top \\ u_2^\top}\right]\left[\matrix{u_1 & u_2}\right]=\left(\left[\matrix{a & b \\ c & d}\right]\left[\matrix{v_1 & v_2}\right]\right)^\top\left[\matrix{a & b \\ c & d}\right]\left[\matrix{v_1 & v_2}\right]=\left[\matrix{1 & 0 \\ 0 & 1 }\right]$
then $\left[\matrix{v_1^\top \\ v_2^\top}\right]\left[\matrix{a & b \\ c & d}\right]^\top\left[\matrix{a & b \\ c & d}\right]\left[\matrix{v_1 & v_2}\right]=\left[\matrix{1 & 0 \\ 0 & 1 }\right]=\left[\matrix{v_1^\top \\ v_2^\top}\right]\left[\matrix{v_1 & v_2}\right]$
Which shows the $(a,b,c,d)$ matrix is orthonormal.