= Solution
Row operations multiply each original tableau equation by an invertible matrix. The <slack variable> block therefore records that matrix and allows the original <payoff matrix> to be recovered without guessing.
From the first tableau, let $X$ be the coefficient block of $x$ and $S$ that of $s$. Since its original equations were $Q^Tx+s=\mathbf1$, we have $Q^T=S^{-1}X$. Here
$$
S^{-1}=\begin{pmatrix}4&0&1\\1&1&1\\3&0&3\end{pmatrix},\qquad X=\begin{pmatrix}0&1&0\\0&0&3\\1&0&1\end{pmatrix},
$$
so
$$
Q^T=\begin{pmatrix}1&4&1\\1&1&4\\3&3&3\end{pmatrix}.
$$
For the second tableau, writing its $y$ and $r$ blocks as $Y,R$, the original equations $Py+r=\mathbf1$ give $P=R^{-1}Y$. We obtain
$$
R^{-1}=\begin{pmatrix}1&1&0\\0&4&0\\0&3&1\end{pmatrix},\qquad Y=\begin{pmatrix}3/4&15/4&0\\1/4&1/4&1\\9/4&9/4&0\end{pmatrix}.
$$
Thus a representative pair of <payoff matrices> is
$$
\boxed{P=\begin{pmatrix}1&4&1\\1&1&4\\3&3&3\end{pmatrix},\qquad Q=P^T=\begin{pmatrix}1&1&3\\4&1&3\\1&4&3\end{pmatrix}.}
$$
Both reconstructed right-hand sides are $\mathbf1$, as a check on the tableau normalization. Independent transformations $P\mapsto aP+b\mathbf1\mathbf1^T$, $Q\mapsto cQ+d\mathbf1\mathbf1^T$, with $a,c>0$, preserve <best responses> and hence identify the same strategic solution up to positive affine payoff transformations.
Directly,
$$
Py'=(1,3,3)^T,\qquad Q^Tx'=Px'=(3,2,3)^T.
$$
The supports of $x'$ and $y'$ lie entirely among their respective best-response coordinates, confirming the <Nash equilibrium> found above. Since this representative is a <symmetric bimatrix game>, swapping the players' strategies preserves the <Nash equilibrium> conditions. Explicitly, $Px'=(3,2,3)^T$ is maximized on the support of $y'$, and $Q^Ty'=Py'=(1,3,3)^T$ is maximized on the support of $x'$. Therefore
$$
\boxed{(y',x')\text{ is also a Nash equilibrium};\quad\text{both players' payoffs are }3\text{ in either profile}.}
$$
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