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Suppose vectors $v_1, v_2$ are orthonormal. Furthermore, suppose that $u_1, u_2$ correspond to two different linear combinations of $v_1, v_2$, and $u_1, u_2$ are also orthonormal.

Could it be deduced that the matrix $U=[u_1 ~u_2]$ is a rotated version of $V=[v_1 ~ v_2]$?

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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.

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Let all of them be $N \times 1$ vectors. Clearly $U=VX$ for some $2\times 2$ matrix $X$. Given $U^{H}U=I$. Hence $(VX)^{H}VX=I$ implies $X^{H}X=I$. In a similar manner, using $V^{H}V=I$, you can prove Note that $XX^{H}=I$. Hence $X$ is a orthogonal matrix. Multiplication by any orthogonal matrix is a rotation.

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