The operator norm is invariant under transposition

Idea. Express the norm as a supremum of a *bilinear* expression in which $x$ and $y$ play symmetric roles, then notice that transposing $A$ simply exchanges them.

Step 1: the bilinear formula. For a vector $v\in\mathbb{R}^n$, $|v|=\sup_{|y|=1}y^Tv$: indeed $y^Tv\le|y||v|=|v|$ by Cauchy-Schwarz, and equality holds for $y=v/|v|$ (any $y$ if $v=0$). Applying this to $v=Ax$,
$$\|A\|=\sup_{|x|=1}|Ax|=\sup_{|x|=1}\sup_{|y|=1}y^TAx=\sup_{|x|=|y|=1}y^TAx.$$

Step 2: transposing. The number $y^TAx$ is a scalar, so it equals its own transpose: $y^TAx=(y^TAx)^T=x^TA^Ty$. Hence
$$\|A\|=\sup_{|x|=|y|=1}x^TA^Ty=\sup_{|y|=|x|=1}x^T A^Ty=\|A^T\|,$$
where in the last step we applied the formula of Step 1 to $A^T$ (with the roles of $x$ and $y$ exchanged).

Remark. An alternative proof: $\|A^T\|^2=\lambda_{\max}(AA^T)$ and $AA^T$, $A^TA$ have the same non-zero eigenvalues (equal to $\sigma_i^2$), so $\|A^T\|=\sigma_1=\|A\|$. The same equality *fails* for the induced norms $\|\cdot\|_1,\|\cdot\|_\infty$: there $\|A^T\|_1=\|A\|_\infty$, which differs in general from $\|A\|_1$.