There is no way to make the operator continuous. Consider the positive semidefinite matrices
$$ A(t) = \pmatrix{\cos^2(t) & \cos(t) \sin(t)\cr \cos(t) \sin(t) & \sin^2(t)\cr}$$
which have eigenvalues (and singular values) $1$ and $0$. The normalized left singular vector for singular value $1$ is $\pm [\cos(t), \sin(t)]$. To make this vector a continuous
function of $t$, you must take either $[\cos(t), \sin(t)]$ for all $t$ or $[-\cos(t), -\sin(t)]$ for all $t$, resulting in the vector at $t=\pi$ being
the negative of the vector at $t=0$. But $A(\pi) = A(0)$, so this can't happen.
First, the difference in the eigenvectors. Let $(\lambda,v)$ be an eigenpair of $A$, i.e., $A v = \lambda v$ and let $\alpha \in \mathbb{C} \setminus \{0\}$. Then
$$A (\alpha v) = \alpha A v = \alpha \lambda v = \lambda (\alpha v).$$
So, $v$ is an eigenvector of $A$ if and only if $\alpha v$ is an eigenvector of $A$. Both are equally "good", unless you desire some additional properties. Note that this works for any $A$, not just $A = C$.
Second, the significance of the left singular vectors is in computing the eigenvalue decomposition in $XX^T$ (in your notation: $X^T = X'$).
Third, a real diagonal matrix is orthogonal if and only if each of its diagonal elements is either $1$ or $-1$. Let us prove this.
Let $D = \mathop{\rm diag}(d_1,\dots,d_n)$. Obviously, $D = D^T$, so
$$D^TD = \mathop{\rm diag}(d_1^2,\dots,d_n^2).$$
So, $D^TD = {\rm I}$ if and only if $d_k^2 = 1$ for all $k$.
For complex matrices (and using complex adjoint $Q^*$ instead of transpose $Q^T$), we get that $|d_k| = 1$ for all $k$.
Best Answer
Yes. This is a consequence of the invariance of the Frobenius norm under orthogonal transformations.
If $Q$ is an orthogonal matrix, then $\| QA \|_{F}= \| A \|_{F}$. Similarly, $\| AQ \|_{F}=\| A \|_{F}$.
Since $U$ and $V$ in the SVD are orthogonal, $\| A \|_{F}=\| \Sigma \|_{F}$.