There are three parts to this question, and I'm not sure how they link together to provide answers.
$A$ is a linear mapping from Euclidean space $X$ to Euclidean space $U$, and the norm $\| \cdot \|$ is the Euclidean norm for matrices.
(i) Show that $\|A^{T}\| = \|A\|$.
(ii) Let $v \in \mathbb{R}^{n}$ be a unit vector, and $\sigma u = Av$, with $\sigma = \|Av\|$. Therefore, $u\in \mathbb{R}^{m}$ is also a unit vector. Does it follow that $\sigma v = A^{T}u$?
(iii) Now if $v$ is as above, but $\sigma = \|A\|$. Show that $\sigma v = A^{T}u$.
Your intuitive idea for (i) is nice, but difficult to implement since $\|A\|$ cannot be nicely written as a function of the entries of $A$. One approach to (i) is as follows. Note that $$ \|A\|^2 = \max_{\|x\| = 1} \|Ax\|^2 = \max_{\|x\| = 1} (Ax)^T(Ax) = \max_{\|x\| = 1}x^T(A^TA)x. $$ By the Rayleigh-Ritz theorem, this is simply the largest eigenvalue of $A^TA$, and the vector at which we attain this maximum is the corresponding eigenvector. Similarly, $\|A^T\|^2$ must be the largest eigenvalue of $AA^T$. Since $A^TA$ and $AA^T$ have the same non-zero eigenvalues (this can be proven in several ways), we conclude that $\|A\| = \|A^T\|$.
(ii): This will not hold in general. For instance, consider $$ A = \pmatrix{1&1\\0&1}, \quad u = v = \pmatrix{1\\0}, \quad \sigma = 1. $$ We indeed have $\sigma u = Av$, but we do not have $A^Tu = \sigma v$.
(iii): The key is to observe that in the case that by the Rayleigh Ritz theorem (as discussed in the first part of this answer), $\|Av\| = \|A\|$ implies that $v$ is an eigenvector of $A^TA$ with $A^TAv = \|A\|^2v$. Thus, with $\sigma = \|A\|$ and the definitions from (ii), we have the following.
If $\sigma = \|A\| = 0$, then it follows that $A = 0$ and the result follows trivially. In the case where $\sigma \neq 0$, we have $$ A^Tu = A^T \left(\frac {Av}{\sigma}\right) = \frac 1{\sigma} A^TA v = \frac 1{\sigma} \sigma^2 v = \sigma v $$ as was desired.