The given circle is ${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}+6\mathrm{y}-5=0$

its center and radius are $\left(2,-3\right)$ and $r=\sqrt{4+9+5}=3\sqrt{2}$

Vertices on square and the circle whose sides are parallel to the coordiante axes are $\left({\mathrm{x}}_1\pm \frac{\mathrm{r}}{\sqrt{2}},{\mathrm{y}}_1\pm \frac{\mathrm{r}}{\sqrt{2}}\right),$

where $\left(x_1,y_1\right)=\left(2,-3\right)$ and radius is $r=3\sqrt{2}$

hence, the vertices are $\left(2\pm 3,-3\pm 3\right)$

it implise all four vertices of square are $\boxed{\left(5,0\right),\left(-1,0\right),\left(5,-6\right),\left(-1,-6\right)}$