At first glance it seems somewhat trivial, but I have some doubts, so I'd like your opinion.
We are given that $\left\|Ax\right\| = \left\|x\right\|, ~ \forall x \in \mathbb{C}^{n}$ and want to show that $\text{cond}\left(A\right) = 1$.
As we know, $\text{cond}\left(A\right) = \left\|A\right\| \left\|A^{-1}\right\|$. Moreover, $$\left\|A\right\| = \sup_{\left\|x\right\| = 1} \left\|Ax\right\|$$ so I choose to work in this direction.
Now, the part which I am uncertain about is this: $$ \left\|Ax\right\| = \left\|x\right\| \Leftrightarrow \left\|A^{-1}Ax\right\| = \left\|A^{-1}x\right\| \Leftrightarrow \left\|x\right\| = \left\|A^{-1}x\right\| $$
so the matrix $A^{-1}$ is also an isometry.
If the previous holds, it follows trivially that $$ \left\|A\right\| = \sup_{\left\|x\right\| = 1} \left\|Ax\right\| = \sup_{\left\|x\right\| = 1} \left\|x\right\| = 1 $$
and the same for $A^{-1}$, hence $\text{cond}\left(A\right) = 1$.
Does this seem correct? Can I apply the multiplication with $A^{-1}$ inside the norm? It doesn't seem to contradict my understanding of linear algebra, but then again, I feel a bit rusty...
If $A$ is an isometry you have $\|Ax\| = \|x\|$, and so $\|A\|=1$.
If $A$ is an isometry, then so it $A^{-1}$. To see this, note that since $\|Ax\| = \|x\|$, $A$ is invertible. Letting $x=A^{-1} y$ gives $\|y\| = \| A^{-1} y \|$. Hence $\|A^{-1}\| =1$.