The if case (where $ABC$ is equilateral) is easy; all one needs to do is to observe the many symmetries an equilateral triangle possesses. It suffices to show the converse: that if $\triangle UVW=\triangle XYZ$ then $ABC$ is equilateral. And to do this, we angle chase. Note that since the two triangles have the same circumradius (since all six points $U,V,W,X,Y,Z$ are concyclic), the problem is reduced to the case where the two triangles are similar. Next, note that \[\angle UWV=\frac12\angle UIV=\frac12(180^\circ-\angle ZAI-\angle ZBI)\] and that \[\angle XZY=\angle XIC=90^\circ-\angle ICX.\] If the two angles $\angle UWV$ and $\angle XZY$ are congruent, then we must have \begin{align*}\tfrac12(180^\circ-\angle ZAI-\angle ZBI)&=90^\circ-\angle ICX\\ 2\angle ICX&=\angle ZAI+\angle ZBI\\ 2\angle C&=\angle A+\angle B=180^\circ-\angle C\\\angle C&=60^\circ.\end{align*} Applying this cyclically, we see that $\triangle ABC$ is equilateral and we're done. $\blacksquare$

it is easy to see that , the angles $\Delta UVW$ are $\frac{\pi+A}{4}$ , $\frac{\pi+B}{4}$ , $\frac{\pi+C}{4}$ and that of $\Delta XYZ$ are $(\pi-A)/2$ ,$(\pi-B)/2$ ,$(\pi-C)/2$ . clearly , if $A=B=C= \pi/3$ , then both of the above mentioned triangles are equilateral and both of them have equal circumradius [which is in fact equal to the inradius of $\Delta ABC$]. hence they are congruent . so , proof of ''if'' part is over. ....................................................................................................................................... for ''only if'' part --- the triangles are congruent . so , the angles of $UVW$ are equal to those of $XYZ$ in some order. so , we get 6 possible cases of equality among angles , namely , 1. $(\pi-A)/2=(\pi+A)/4$ ,$ (\pi-B)/2=(\pi+B)/4$ 2. $(\pi-A)/2=(\pi+A)/4$ ,$(\pi-B)/2=\frac{\pi+C}{4}$ 3. $(\pi-A)/2=(\pi+B)/4$ ,$(\pi-B)/2=\frac{\pi+C}{4}$ 4. $(\pi-A)/2=(\pi+B)/4$ , $(\pi-B)/2=(\pi+A)/4$ 5. $(\pi-A)/2=\frac{\pi+C}{4}$ ,$(\pi-B)/2=(\pi+B)/4$ 6. $(\pi-A)/2=\frac{\pi+C}{4}$ , $(\pi-B)/2=(\pi+A)/4$ in each of these cases , when solved , we get $A=B=C=\pi/3$ . hence done.

Starting off the day with a simple problem. The reverse direction is easy. Just note that $\angle UWV=\frac{1}{2}(UIW)=\frac{1}{2}AIC$ which is known. Also $\angle XZY$ is known,hence after some simple steps we are done.

If the triangle is equilateral then we have that , $XYZ$ is also the orthic triangle of $\triangle{ABC}$ and hence we have $I$ as the incenter of $\triangle{XYZ}$ and since $\triangle{XYZ}$ is equilateral we have $\triangle{UVW}$ to be equilateral and since all of them have the same circumradius we have $\triangle{XYZ}\cong \triangle{UVW}$ now we deal with the case when triangles are congruent we $\textbf{\textcolor{red}{claim:-}}$ $I$ is the circumcenter of $\triangle{XYZ}$ and $\triangle{UWV}$ $\textbf{\textcolor{blue}{proof:-}}$ since we have $\angle{ZXY}=\frac{180^{\circ}-ZAY}{2}$ and $\angle{ZIY}=180^{\circ}-\angle{ZAY}$ we have $I$ as the circumcenter of $\triangle{XYZ}$ also since points $U,V,W,X,Y,Z$ are concyclic they have same circumcenter so $\triangle{UWV}$ has also circumcenter $I$ $\square$ now we since we have $\angle{UIV}=\frac{\angle{UWV}}{2}=\frac{1}{2}\cdot \left(180^{\circ}-\left(\angle{BAI}+\angle{IBA}\right)\right)=\frac{1}{2}\cdot \left(90^{\circ}+\frac{\angle{ACB}}{2}\right)$ and also this angle is equal $\angle{YZX}=\frac{180^{\circ}-\angle{ACB}}{2}$ we get $180^{\circ}-\angle{ACB}=90^{\circ}+\frac{\angle{ACB}}{2}\implies \angle{ACB}=60^{\circ}$ so from similar argument we can prove that $\angle{ABC}=\angle{BAC}=60^{\circ}$ hence we have $\triangle{ABC}$ as equilateral triangle and we are done $\blacksquare$