Suppose $e_1,e_2,\cdots,e_n$ and $f_1,f_2,\cdots,f_n$ are orthonormal bases of vector space $V$ and that the singular value decomposition of a linear operator $L:V\to V$ is $$ Lv=\sum_{i=1}^{n}s_i\langle v,e_i\rangle f_i. $$ It could be shown that $$ L^*v=\sum_{i=1}^{n}s_i\langle v,f_i\rangle e_i. $$ And combining these two equations gives us that $$ L^*Lv=\sum_{i=1}^{n}s_i^2\langle v,e_i\rangle e_i,LL^*v=\sum_{i=1}^{n}s_i^2\langle v,f_i\rangle f_i. $$ We could also know, from polar decomposition theorem, that there exists an isometric isomorphism $S:V\to V$ such that $L=S\sqrt{L^*L}$. Then $e_1,e_2,\cdots,e_n$ are orthonormal eigenvectors of $\sqrt{L^*L}$ and $f_1,f_2,\cdots,f_n$ are orthonormal eigenvectors of $\sqrt{LL^*}$ and $f_j=S(e_j),\forall j=1,2,\cdots,n$.
Now suppose $L$ is a normal operator (i.e. $L^*L=LL^*$), then is it true that $f_j=t_je_j$ i.e. $S:e_j\mapsto t_je_j$ for all $j$ where $t_j\in\mathbb{C}$ for all $j$? The converse of this proposition is perfectly true, as $L^*L=LL^*$ only requires $|t_j|=1$, which is true since $S$ is an isometry.