I assume $\sigma_1\neq 0$, which means that $\Sigma$ is invertible. Let $A=Q\Sigma^{-1}.$ Then $Q=A\Sigma$ and $Q^T=\Sigma A^T.$
We want $Q^T \Sigma$ to be symmetric, which means $Q^T\Sigma = \Sigma Q$ or
$\Sigma A^T \Sigma = \Sigma A\Sigma.$ If we multiply this with $\Sigma^{-1}$ from both sides, we get $A=A^T,$ so $A$ is symmetric.
$Q$ is orthogonal, which means $Q^TQ=I$ or
$$
\Sigma A^2 \Sigma = I
$$
or
$$
A^2 = \Sigma^{-2}
$$
So we are looking for a square root of $\Sigma^{-2}$ and the problem boils down to the question if $\Sigma^{-1}$ is the only valid choice.
We must consider the case that $\Sigma$ has eigenvalues with multiplicity of more than $1.$
Let $\sigma_{r_i} = \sigma_{r_i+1} = \ldots = \sigma_{r_{i+1}-1}$ for $i=1,\ldots,m$ and $r_1=1,$ $r_2=2$ and $r_{m+1}=n+1.$ Furthermore, $\sigma_{r_i}<\sigma_{r_{i+1}}$ for $i=1,\ldots,m-1.$ Then each square root of $\Sigma^{-2}$ can be written as follows
$$
A = \begin{pmatrix}
\sigma_{r_1}^{-1} B_1 & & & & 0 \\
& \sigma_{r_2}^{-1} B_2 & & & \\
& & \sigma_{r_3}^{-1} B_3 & & \\
& & & \ddots & \\
0 & & & & \sigma_{r_m}^{-1} B_m
\end{pmatrix}
\;\;,\;\;
B_i^2 = I\;\;\mbox{for}\;\; i=1,\ldots,m
$$
where $B_i$ are blocks of size $(r_{i+1}-r_i)\times (r_{i+1}-r_i).$ (The proof is given below)
Then
$$
Q = \begin{pmatrix}
B_1 & & & & 0 \\
& B_2 & & & \\
& & B_3 & & \\
& & & \ddots & \\
0 & & & & B_m
\end{pmatrix}
$$
The $B_i$ are symmetric. $B_i^T$ is the inverse of $B_i$ because of the orthogonality of $Q$, and $B_i$ is also the inverse of $B_i$, because of the property $B_i^2=I.$ Therefore $B_i^T=B_i$ and
$$
Q^T\Sigma = \begin{pmatrix}
\sigma_{r_1}B_1 & & & & 0 \\
& \sigma_{r_2}B_2 & & & \\
& & \sigma_{r_3}B_3 & & \\
& & & \ddots & \\
0 & & & & \sigma_{r_m}B_m
\end{pmatrix}
$$
We want $Q^T\Sigma$ to have the same eigenvalues as $\Sigma,$ which in turn means that $\sigma_{r_i}B_i$ has $\sigma_{r_i}$ as its only eigenvalue. A symmetric matrix with only one eigenvalue must be a scalar multiple of the identity matrix. Therefore, $B_i = I$ for $i,\ldots,m,$ which completes the proof.
Proof sketch for $\sigma_1=0$
If $\sigma_1=0,$ it can easily be shown that $Q_{11}\in\{-1,1\}$ and $Q_{1j}=Q_{j1}=0$ for $j=2,\ldots,n.$ This can be concluded from the symmetry of $Q^T\Sigma$ and from the orthogonality of $Q.$
This means that we can follow the argument from the first part of the proof, but consider only the subspace that is orthogonal to $e_1.$ Basically, this means that we ignore the first row and first column of all $n\times n$ matrices. In the end, we have to decide if $Q_{11}=1$ or $Q_{11}=-1.$ As $Q\in \mathrm{SO}(n)$ and $B_i=I$ for $i=2,\ldots,m,$ we can conclude $Q_{11}=1.$
Diagonalizable square roots of diagonal matrices
Let $A$ be diagonalizable and $A^2$ diagonal. Without loss of generality, the diagonal elements of $A^2$ are sorted in ascending order.
Let $0\leq\lambda_1 < \lambda_2 < \ldots < \lambda_m$ such that the eigenvalues of $A$ form a (not necessarily strict)
subset of $\{\lambda_1,\;-\lambda_1,\;\lambda_2,\;-\lambda_2,\;\ldots,\;\lambda_m,\;-\lambda_m\}.$
Let $t_i^{+}$ be the algebraic and geometric multiplicity of $\lambda_i$ and
$t_i^{-}$ the algebraic and geometric multiplicity of $-\lambda_i$ within the matrix $A$ (we set $t_1^{-}=0$ if $\lambda_1=0.$)
Let $r_1=1$ and $r_{i+1} = r_i + t_i^{+}+ t_i^{-}.$
If $Av = \lambda v$ and $Aw = -\lambda w,$ then $A^2 (v+w) = A^2 v + A^2 w =\lambda^2 v + (-\lambda)^2 w = \lambda^2 (v+w).$
This means that the eigenspace of $A^2$ with respect to the eigenvalue $\lambda^2$ is the direct sum of the
eigenspaces of $A$ with respect to the eigenvalues $\lambda$ and $-\lambda.$
As $A$ is diagonalizable, the direct sum of the eigenspaces $E_{A,\lambda_1},$ $E_{A,-\lambda_1}$,
$E_{A,\lambda_2},$ $E_{A,-\lambda_2},\ldots$,
$E_{A,\lambda_m},$ $E_{A,-\lambda_m}$,
forms the complete vector space $\mathbb{R}^n.$
This means that each of the eigenspaces of $A^2$ can be written as
$E_{A,\lambda_i} \oplus E_{A,-\lambda_i}.$
In a manner of speaking, there is no room for other eigenspaces than those.
We know the eigenspaces of $A^2,$ because $A^2$ is diagonal. We have
\begin{eqnarray*}
E_{A^2,\lambda_1^2} & = & E_{A,\lambda_1} \oplus E_{A,-\lambda_1} = \mathrm{span}\{e_{r_1},\ldots,e_{r_2-1}\} \\
& \vdots & \\
E_{A^2,\lambda_m^2} & = & E_{A,\lambda_m} \oplus E_{A,-\lambda_m} = \mathrm{span}\{e_{r_m},\ldots,e_{r_{m+1}-1}\}
\end{eqnarray*}
with the standard basis $e_1,\ldots,e_n.$
Now it is clear that $A$ can be diagonalized by means of a block matrix, because each $E_{A,\lambda_i} \oplus E_{A,-\lambda_i}$ is spanned by the
related elements of the standard basis.
$$
A=
\begin{pmatrix}
S_1 & & 0 \\
& \ddots & \\
0 & & S_m
\end{pmatrix}
\begin{pmatrix}
\lambda_1 I_{t_1^{+}} & & & & 0 \\
& -\lambda_1 I_{t_1^{-}} & & & \\
& & \ddots & & \\
& & & \lambda_m I_{t_m^{+}} & \\
0 & & & & -\lambda_m I_{t_m^{-}}
\end{pmatrix}
\begin{pmatrix}
S_1 & & 0 \\
& \ddots & \\
0 & & S_m
\end{pmatrix}
^{-1}
$$
From this, by simply processing the matrix multiplication, we can conclude that $A$ itself is also a block matrix of the same sort, i.e.
$$
A = \begin{pmatrix}
A_1 & & 0 \\
& \ddots & \\
0 & & A_m
\end{pmatrix}
$$
with
$$
A_i = S_i\,\begin{pmatrix}
\lambda_i I_{t_i^{+}} & \\
& -\lambda_i I_{t_i^{-}} \\
\end{pmatrix}
\,
S_i^{-1}
$$
Now we only have to show that $A_i = \lambda_i B_i$ with $B_i^2=I.$
Let $T_i=S_i^{-1}$.
Let $S_i^{+}$ be the $(t_i^{+}+t_i^{-})\times t_i^{+}$ matrix that is formed by the first $t_i^{+}$ columns of $S_i$ and
$S_i^{-}$ the $(t_i^{+}+t_i^{-})\times t_i^{-}$ matrix that is formed by the last $t_i^{-}$ columns of $S_i.$
Let $T_i^{+}$ be the $t_i^{+}\times (t_i^{+}+t_i^{-})$ matrix that is formed by the first $t_i^{+}$ rows of $T_i$ and
$T_i^{-}$ the $t_i^{-}\times (t_i^{+}+t_i^{-})$ matrix that is formed by the last $t_i^{-}$ rows of $T_i.$
Then $T_i^{+}S_i^{+}=I,\;\;T_i^{-}S_i^{-}=I,\;\;T_i^{+}S_i^{-}=0,\;\;T_i^{-}S_i^{+}=0$.
$$
A_i = S_i^{+}\lambda_i T_i^{+} + S_i^{-}(-\lambda_i) T_i^{-} = \lambda_i \left( S_i^{+}T_i^{+} - S_i^{-}T_i^{-}\right)
$$
Let $B_i = S_i^{+}T_i^{+} - S_i^{-}T_i^{-}.$ Then
\begin{eqnarray*}
B_i^2 & = & \left( S_i^{+}T_i^{+} - S_i^{-}T_i^{-}\right)\left( S_i^{+}T_i^{+} - S_i^{-}T_i^{-}\right) \\
& =& S_i^{+}T_i^{+}S_i^{+}T_i^{+}-S_i^{+}T_i^{+}S_i^{-}T_i^{-}-S_i^{-}T_i^{-}S_i^{+}T_i^{+}+S_i^{-}T_i^{-}S_i^{-}T_i^{-} \\
& =& S_i^{+}\cdot I\cdot T_i^{+}-S_i^{+}\cdot 0 \cdot T_i^{-}-S_i^{-}\cdot 0 \cdot T_i^{+}+S_i^{-}\cdot I\cdot T_i^{-} \\
& =& S_i^{+}T_i^{+}+S_i^{-}T_i^{-} \\
& =&
\begin{pmatrix}
& & \\
S_i^{+} & & S_i^{-} \\
& &
\end{pmatrix}
\begin{pmatrix}
& T_i^{+} & \\
& & \\
& T_i^{-} &
\end{pmatrix}
=S_iT_i = I
\end{eqnarray*}
Best Answer
Notation $$(\delta_i,\sigma_j)$$ should be interpreted as the value of Dirac $\delta_i$ acting on $\sigma_j$ (see for example here), i.e.,
$$\delta_i(\sigma_j)=\begin{cases}1 & \text{if} \ \sigma_j=i \iff \sigma(j)=i\\0 & \text{otherwise}\end{cases}$$
giving indeed exactly one "$1$" on every line.
For your second part: on inspection of a permutation matrix, which is known to have one "1" per line and per column (it is their definition), on each line "i", there is a unique "one" situated on a certain column $j$ determines the image $\sigma(j)=i$. When all lines have been processed, $\sigma$ is fully determined. But $\sigma$ has to be proven injective. This is where the other unicity intervenes: no chance that $\sigma(j_1)=\sigma(j_2)$ for $j_1 \ne j_2$ because there is a unique "one" per column.
The result is indeed the classical permutation matrix associated to $\sigma$...
Remark: notation $(u,v)$ is indeed puzzling ; it is usually used for dot products, but sometimes as well for the "action" of the first argument onto the second: $(f,x):=f(x)$, mainly in the context of duality (like in the reference I gave above).