Let $\,\boldsymbol{x}(t)$, $\boldsymbol{y}(t)$, $\boldsymbol{z}(t)$ and
$\boldsymbol{w}(t)$ be the vectors describing the positions of the four ants,
and let
$$
\boldsymbol{A} \,=\, (1,1), \,\, \boldsymbol{B} \,=\, (-1,1), \,\,
\boldsymbol{\varGamma} \,=\, (-1,-1) \,\,\, \text{and} \,\,\,
\boldsymbol{\varDelta} \,=\, (1,-1),
$$
be their initial positions. As the first ant chases the second one, the velocity $\,\boldsymbol{x}'(t)\,$ of the first ant will have the same direction with the vector $\,\boldsymbol{y}(t)-\boldsymbol{x}(t),\,$ which connects the two ants, i.e., $$\,\boldsymbol{x}'(t)\,\varpropto\,\boldsymbol{y}(t)-\boldsymbol{x}(t),$$
and as its velocity is of unit norm, $\,\|\boldsymbol{x}'(t)\|=1,$ where $\,\|\cdot \|\,$ is the Eucidean norm in $\mathbb R^2$, then
$$\,\boldsymbol{x}^\prime \,=\, \frac{\boldsymbol{y-x}}{\|\, \boldsymbol{y-x} \|}.$$
Similarly we obtain the equations of motion of the remaining three ants. We thus obtain the following system of ODEs:
$$
\boldsymbol{x}^\prime \,=\, \frac{\boldsymbol{y-x}}{\| \boldsymbol{y-x} \|}, \,\,\,
\boldsymbol{y}^\prime \,=\, \frac{\boldsymbol{z-y}}{\| \boldsymbol{z-y} \|}, \,\,\,
\boldsymbol{z}^\prime \,=\, \frac{\boldsymbol{w-z}}{\| \boldsymbol{w-z} \|}, \,\,\,
\boldsymbol{w}^\prime \,=\, \frac{\boldsymbol{x-w}}{\| \boldsymbol{x-w} \|},
\tag{i}
$$
with initial conditions
\begin{equation}
\boldsymbol{x}(0)=\boldsymbol{A},\,\,\,\boldsymbol{y}(0)=\boldsymbol{B},\,\,\,
\boldsymbol{z}(0)=\boldsymbol{\varGamma},\,\,\,\boldsymbol{w}(0)=\boldsymbol{\varDelta}. \tag{ii}
\end{equation}
Equations $(i)$ and $(ii)$ form an initial value problem (IVP) of a nonlinear $8\!\times\!8$ system of ODEs. If the vectors $\boldsymbol{x}(t)$,
$\boldsymbol{y}(t)$, $\boldsymbol{z}(t)$ and $\boldsymbol{w}(t)$
satisfy our IVP and
\begin{equation*}
U \,=\, \left(\! \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}
\!\right)\!,
\end{equation*}
the orthogonal matrix, rotation by $\pi/2$,
then so do the vectors $U\boldsymbol{w}(t)$,
$U\boldsymbol{x}(t)$, $U\boldsymbol{y}(t)$ and $U\boldsymbol{z}(t)$. Global uniqueness of the solutions of our IVP implies that
\begin{equation}
\boldsymbol{y}\,=\,U\boldsymbol{x},\,\,\,\, \boldsymbol{z} \,=\,
U\boldsymbol{y},\,\,\,\, \boldsymbol{w} \,=\, U\boldsymbol{z}
\,\,\,\,\,\,\text{and}\,\,\,\,\,\, \boldsymbol{x}\,=\,U\boldsymbol{w}.
\end{equation}
This symmetry reduces $\mathrm{(i)-(ii)}$, from an IVP of an $8\times 8$ system, to one of a $2\times 2$ system
$$
\boldsymbol{x}^\prime\,=\, \frac{{U}\boldsymbol{x}-\boldsymbol{x}}
{\|U\boldsymbol{x}-\boldsymbol{x}\|}, \quad \boldsymbol{x}(0) \,=\,\boldsymbol{A}\,=\,(1,1). \tag{1}
$$
Let $\boldsymbol{\varphi}$ be the solution of (1).
Then $\boldsymbol{\varphi}$ shall also satisfy
\begin{equation}
\boldsymbol{\varphi}^\prime(t) \, = \, \alpha(t)
\bigl(U\boldsymbol{\varphi}(t)-\boldsymbol{\varphi}(t)\bigr) \, = \,
\beta^\prime(t)(U-{\mathcal I})\boldsymbol{\varphi}(t), \tag{2}
\end{equation}
where
\begin{equation}
\alpha(t)\,=\, \| U\boldsymbol{\varphi}(t)-\boldsymbol{\varphi}(t) \|^{-1} \qquad\text{and}\qquad \beta(t)\,=\,\int_0^t\alpha(s)\,ds.
\end{equation}
Hence
\begin{equation}
\boldsymbol{\varphi}(t)\,=\,
\mathrm{e}^{\beta(t){(U-{\mathcal I})}}\boldsymbol{\varphi}(0),
\end{equation}
where ${\mathcal I}$ is the unit $2\!\times\! 2$ matrix.
Here we used the fact that if $\,\boldsymbol{w}'=f(t)B\boldsymbol{w},\,$
where $\boldsymbol{w}$ is an $n-$vector, $f$ a scalar function and $B$ a constant $n\times n$ matrix, then
$\,\boldsymbol{w}(t) =\mathrm{e}^{(\int_0^t f(s)\,ds)B}\boldsymbol{w}(0).$
We also have that
\begin{equation}
\mathrm{e}^{t(U-{\mathcal I})}
\,=\, \mathrm{e}^{-t{\mathcal I}}\mathrm{e}^{tU}
\,=\, \mathrm{e}^{-t}\mathrm{e}^{tU}
\,=\,
\left(\!
\begin{array}{rr} \mathrm{e}^{-t}\cos t & -\mathrm{e}^{-t} \sin t
\\ \mathrm{e}^{-t}\sin t & \mathrm{e}^{-t}\cos t
\end{array} \!\right)\!,
\end{equation}
and consequently
\begin{align*}
\boldsymbol{\varphi}(t) =\mathrm{e}^{-\beta(t)}
\left(\! \begin{array}{rr}
\cos \beta(t) & -\sin \beta(t) \\ \sin \beta(t) & \cos \beta(t)
\end{array} \!\right)
\left(\! \begin{array}{r} 1 \\ 1 \end{array} \!\right)
=\mathrm{e}^{-\beta(t)}
\big(\cos\beta(t)-\sin\beta(t),\cos\beta(t)+\sin\beta(t)\big).
\end{align*}
Also
\begin{equation}
(U-{\mathcal I}) \boldsymbol{\varphi}(t)=
\left(\!\!\begin{array}{rr}
-1 & -1 \\ 1 & -1
\end{array}\!\right) \boldsymbol{\varphi}(t) \,=\,
-2\mathrm{e}^{-\beta(t)} \bigl( \cos
\beta(t), \sin\beta(t)\bigr),
\end{equation}
and therefore,
\begin{equation}
\big\| (U-{\mathcal I}) \, \boldsymbol{\varphi}(t) \big\|
= 2\mathrm{e}^{-\beta(t)}. \tag{3}
\end{equation}
Since $\,\| \boldsymbol{\varphi}^\prime(t) \| =1$, then
combining (2) and (3) we obtain
\begin{align}
1=\big\| \boldsymbol{\varphi}^\prime(t) \big\|
=\big\|\beta^\prime(t)(U-{\mathcal I})\boldsymbol{\varphi}(t)\big\|
=2\,\mathrm{e}^{-\beta(t)}\beta'(t).
\end{align}
Integrating the above in the interval $[0,t]$ we obtain that
\begin{equation*}
\mathrm{e}^{-\beta(t)}\,=\,1-\frac{t}{2},
\end{equation*}
since $\,\beta(0)=0$. Thus,
\begin{equation*}
\beta(t)\,=\,-\log\left(1-\frac{t}{2}\right).
\end{equation*}
Hence
\begin{equation}
\big\| \boldsymbol{\varphi}(t)-\boldsymbol{\chi}(t) \big\| \,=\, \big\| (U-{\mathcal I})\,\boldsymbol{\varphi}(t) \big\|=2\mathrm{e}^{-\beta(t)} =2-t.
\end{equation}
Thus, the four ants, at any given moment $t$, before converging to the centre
of the square, they lie on the vertices on a rotating and shrinking square of side
$$
\mu(t)=\big\| \boldsymbol{\varphi}(t)-\boldsymbol{\chi}(t) \big\|=2-t,
$$
where $\,\boldsymbol{\chi}(t)$ is the position of the next ant.
Note that the shrinking rate is constant ($d\ell/dt=1$), while the rotation is described by the angle
$$
\beta(t)\,=\,-\log\left(1-\frac{t}{2}\right),
$$
tends to infinity, as $t$ tends to the time of convergence
$T=2$. Until the convergence in the center, each ant will have covered distance
\begin{equation*}
s = \int_0^{2} \! \| \boldsymbol{x}^\prime(t) \|\,dt =2,
\end{equation*}
i.e., each ant covers distance equal to the side of the square, before they all meet in the center of the square.
We also obtain the trajectories of the ants
\begin{align*}
\boldsymbol{\varphi}(t)&=\frac{2-t}{\sqrt{2}}
\bigl( \cos(\beta(t)-\pi/4),\sin(\beta(t)-\pi/4) \bigr) \\
&=\frac{2-t}{\sqrt{2}}
\bigg(
\cos\Big(\log\Big(1-\frac t2\Big)+\frac{\pi}{4}\Big),
-\sin\Big(\log\Big(1-\frac t2\Big)+\frac{\pi}{4}\Big)
\bigg).
\end{align*}
Observe that the ant will have made infinitely many rounds around the center of the square before they converge there.
