For a square matrix A, I want to show that $$ | \mathbf{A}^T| = |\mathbf{A}| $$
Proof:
Let the same permutation that changes $ \varphi(j_1,...,j_n) $ into $ \varphi(1,...,n) $, change $ \varphi(1,...,n) $ into $ \varphi(i_1,...,i_n) $. Then, noticing that $ \varphi(1,...,n) $ = $ 0 $, we obtain
$$ |\mathbf{A}| = \sum_{(j_1,...,j_n)} (-1)^{\varphi(j_1,...,j_n)}a_{1j_1}a_{2j_2}...a_{nj_n} \quad ...(1) \quad \text{by def. laplace expansion} $$ $$ = \sum_{(j_1,...,j_n)} (-1)^{\varphi(1,...,n)}(-1)^{\varphi(j_1,...,j_n)}a_{1j_1}a_{2j_2}...a_{nj_n} \quad ...(2)\quad \text{by obs.} $$
Now comes the problem. My book then proceeds to this equality which I need to unpack.
$$ = \sum_{(1,...,n)} (-1)^{\varphi(i_1,...,i_n)}(-1)^{\varphi(1,...,n)}a_{i_11}a_{i_2 2}...a_{i_n n} $$
In essence, I don't quite get why one is allowed to change the indices in the summation and multiplication. From the first statement, I gather that this is a constructive proof. Nevertheless, I don't follow how the $\varphi$ can be interchanged. Intuitively, I imagine that they represent the same amount of transpositions (even or odd).
I think that keeping track of these $i_r$'s and and $j_s$'s is very confusing, so I will write what I think is the same proof in a different notation. I use $\pi$ as a general permutation of the set $\{1,2,3,\dots,n\}$. In your question, this is the permutation taking each $r$ to $i_r$, or equivalently taking each $j_s$ to $s$.
I would use $\mathop{sgn}(\pi)$ to denote the sign of the permutation $\pi$. (This is your $\phi$ but I don't believe that your book is using it consistently).
Then $$ \det A=\sum_{\pi} (-1)^{\mathop{sgn}(\pi)} a_{1,\pi(1)}\dots a_{n,\pi(n)} $$ is the Laplace expansion.
We can rewrite this by shuffling the terms as
$$ \sum_{\pi} (-1)^{\mathop{sgn}(\pi)} a_{\pi^{-1}(1),1}\dots a_{\pi^{-1}(n),n}. $$
Now it is easy to see that $\pi$ is odd if and only if $\pi^{-1}$ is odd, so we have in fact $$ \sum_{\pi} (-1)^{\mathop{sgn}(\pi^{-1})} a_{\pi^{-1}(1),1}\dots a_{\pi^{-1}(n),n}. $$
Now as $\pi$ runs through the set of all permutations so does $\pi^{-1}$, so what we really have is
$$ \sum_{\pi^{-1}} (-1)^{\mathop{sgn}(\pi^{-1})} a_{\pi^{-1}(1),1}\dots a_{\pi^{-1}(n),n}. $$
Writing $\rho$ for $\pi^{-1}$ this is
$$ \sum_{\rho} (-1)^{\mathop{sgn}(\rho)} a_{\rho(1),1}\dots a_{\rho(n),n} $$ which is the Laplace expansion of $\det A^{T}$ and we are done.