$\DeclareMathOperator\Sym{Sym}\DeclareMathOperator\Coh{Coh}$I have broken my question into a few sections for clarity and to provide sufficient context to the problem. I apologize for the length. The question is asked in the last section titled "Question".

**Motivation**

The Graph Reconstruction Conjecture states that any two simple graphs with equal decks (hypomorphic graphs) are isomorphic for graphs with $3$ or more vertices. An alternative statement of this conjecture is that for any deck of a simple graph, all corresponding "reconstructions" are isomorphic.

To make "reconstruction" precise, for a deck $D(G)$ of a simple graph of order $n$, we may choose an enumeration of each subgraph so that $D(G)=\{G_1,...,G_n\}$. For each subgraph, we may choose a bijection $\phi_i:V(G_i)\to [n]\setminus\{i\}$ where $V(G_i)$ is the vertex set of $G_i$. Moreover, we must impose that if $\{\phi_i(x),\phi_i(y)\}\in E(G_i)$ then $\{\phi_i(x),\phi_i(y)\}\in E(G_j)$ if $j\ne \phi_i(x)$ and $j\ne\phi_i(y)$. In words, if our choice of labeling on subgraph $i$ implies that if $\{a,b\}$ is an edge then $\{a,b\}$ must be an edge in any other subgraph so long as $a$ and $b$ are both present.

**An Algebraic View of Reconstruction**
Let $D(G)$ be an unlabelled deck of some simple graph $G$ of order $n$. Choose an enumeration of the subgraphs such that $D(G)=\{G_1,...,G_n\}$. For each $G_i$ we may associate an adjacency matrix of size $(n-1)\times(n-1)$ and we would like to append a row and column of all zeros to these matrices so that they have dimension $n\times n$. We will refer to these matrices as $A_i$ for each $i\in \{1,...,n\}$.

Define,

$$\Psi(S)=\displaystyle\frac{1}{n-2} \displaystyle\sum_{i=1}^{n} S_i A_i S_i^T$$

where $S=(S_1,...,S_n)\in \Sym(n)^n$ and $\Sym(n)$ is the group of $n\times n$ permutation matrices. It can be easily demonstrated that,

- $\mathrm{tr}(\Psi(S))=0$
- $\Psi(S)$ is symmetric

The motivation of $\Psi$ is to encapsulate the reconstruction problem as posed in the motivation section above - the choice of labeling is equivalent to choosing some $n$-tuple of permutations on the "extended" subgraph adjaceny matrices (i.e. $A_i$). We say that $S$ is "coherent" if $\Psi(S)$ is a valid adjacency matrix. Moreover, we denote the set of all coherent values of $S$ as,

$$\Coh(\Psi) = \{S\in \Sym(n)^n \mid \Psi(S)\circ \Psi(S)=\Psi(S)\}$$

where $\circ$ denotes the Hadamard product. It is easy to show that this condition is equivalent to asserting that each entry of $\Psi(S)$ be either $0$ or $1$. This fact coupled with the symmetry and $0$ trace conditions above imply that $\Psi(S)$ would be a valid adjacency matrix.

The motivation of $\Coh(\Psi)$ is to algebraically encapsulate the set of all "coherent" reconstructions from the deck $D(G)$.

* Question*
Is the following statement equivalent to the graph reconstruction conjecture?

For any $S_1, S_2\in \Coh(\Psi)$ there exists $Q\in \Sym(n)$ such that,

$\Psi(S_1) = Q \Psi(S_2) Q^T$

**Additional Context**
For additional context, if this reformulation is in fact valid. I am looking to proceed as follows: Let $S_1, S_2\in \Coh(\Psi)$ and suppose $\Psi(S_1)$, $\Psi(S_2)$ are counterexamples to the GRC. Since these matrices are real and symmetric $\Lambda_1 = Q_1\Psi(S_1)Q_1^T$ and $\Lambda_2 = Q_2\Psi(S_2)Q_2^T$ for orthogonal matrices $Q_1$ and $Q_2$ and diagonal matrices of eigenvalues $\Lambda_1$, $\Lambda_2$. Tutte proved that the characteristic polynomial is reconstructable and thus a counterexample must be cospectral. Thus there exists $P\in \Sym(n)$ such that $\Lambda_1 = P\Lambda_2P^T$. Some algebra implies that,

$$\Psi(S_1) = (Q_1^TPQ_2)\Psi(S_2)(Q_2^TPQ_1)$$

It is clear that $(Q_1^TPQ_2)$ must be orthogonal. If we can show that $(Q_1^TPQ_2)\in \Sym(n)$ then this implies the GRC (of course assuming I have not made a mistake in reformulating).