Fixed circumcircles $(X),(Y),(Z)$ of $\triangle A'BC , \triangle C'AB , \triangle B'CA$ meet at the Miquel point of $\triangle A'B'C' \cup ABC,$ i.e. the 1st Fermat point $F$ of $\triangle ABC.$ Let $T$ be the midpoint of $B'C',$ running on midcircle $(K)$ of $(Y),(Z),$ and let $L$ be the 2nd intersection of $A'T$ with $(X);$ midpoint of its arc $BFC.$ Since $AF$ bisects $\angle BFC,$ it follows that $FL$ is the external bisector of $\angle BFC$ $\Longrightarrow$ $\angle AFL=90^{\circ}$ $\Longrightarrow$ $L \in (K).$ Let $G$ and $U$ be the centroids of $\triangle ABC$ and $\triangle A'B'C'.$ Since $G$ is also the centroid of $\triangle XYZ,$ then $\frac{_{\overline{GX}}}{^{\overline{GK}}}=-2 \ (*).$ But $\frac{_{\overline{UA'} \cdot \overline{UL}}}{^{\overline{UL} \cdot \overline{UT}}}=-2$ $\Longrightarrow$ powers of $U$ WRT circles $(X),(K)$ are in the same ratio $\Longrightarrow$ $U$ is on circle $\Omega$ coaxal with $(X),(K),$ but from $(*)$ we deduce that $G$ is the center of $\Omega.$ Therefore, locus of $U$ is the circle $\Omega$ centered at the centroid $G$ of $\triangle ABC$ and passing through its 1st Fermat point $F.$

Hint : take midpoints of arc's AB , BC , CA of (ABC') , (BCA') , (CAB')

Another proof, following skytin's idea: Circumcircles $(X),(Y),(Z)$ of $\triangle A'BC,$ $\triangle B'CA$ and $\triangle C'AB$ concur at the Fermat point $F$ of $\triangle ABC.$ Since $U$ is the incenter of $\triangle A'B'C',$ then $UA',UB',UC'$ cut $(X),(Y),(Z)$ again at the midpoints $A_0,B_0,C_0$ of its arcs $FBC,FCA,FAB,$ these are none other than the reflections of $X,Y,Z$ on $BC,CA,AB.$ Thus $\triangle A_0B_0C_0$ is the inner equilateral Napoleon triangle of $\triangle ABC,$ whose center is the centroid $G$ of $\triangle ABC.$ Its circumcircle passes through $F$ and $U,$ since $\angle (FC_0,FB_0)=60^{\circ}$ and $\angle (UC_0,UB_0)=60^{\circ}$ $\pmod\pi.$ The conclusion follows.

A more generalized decision: Let $K$ be an arbitrary point in the plane.Draw circles $\odot(BKC),\odot(CKA),\odot(AKB)$. We'll prove that the locus of the centroid $G$ of triangles like $\triangle{MNP}$ such that $P \in \odot(BKC),N \in \odot(CKA),M \in \odot(AKB)$ and $MN,MP,NP$ passes through $A,B,C$ respectively, is a circle centered at the centroid $L$ of triangle $\triangle{A'B'C'}$ passing through $K$ where $A',B',C'$ are the circumcircles of $\odot(BKC),\odot(CKA),\odot(AKB)$ respectively. $\rightarrow \vec{LG} = \frac{1}{3}.(\vec{MC'}+\vec{NB'}+\vec{PA'})$. $\rightarrow$ Now let $M$ traverses a distance equal to $\alpha$ on the perimeter of $\odot(AKB)$ and makes the point $M'$ ($\widehat{M'C'M}=2\alpha$). WLOG suppose the movement is clockwise; Then if we call the intersection points of $M'A$ and $M'B$ with $\odot(CKA)$ and $\odot(BKC)$ as $N',P'$ respectively, we obviously have $\widehat{M'C'M}=\widehat{N'B'N}=\widehat{P'A'P}=2\alpha \Rightarrow$ If $G'$ be the centroid of $\triangle{M'N'P'}$ then $\vec{GG'}$ is the rotatated vector of $\vec{LG}$ with angle $2\alpha$(Note that $N'$ and $P'$ are also traversing clockwise). So the length of $\vec{LG}$ is constant $\Longrightarrow$ The locus of $G$ is a circle centered at the centroid $L$ of triangle $\triangle{A'B'C'}$ passing through $K \blacksquare$.

wiseman wrote: A more generalized decision: Let $K$ be an arbitrary point in the plane.Draw circles $\odot(BKC),\odot(CKA),\odot(AKB)$. We'll prove that the locus of the centroid $G$ of triangles like $\triangle{MNP}$ such that $P \in \odot(BKC),N \in \odot(CKA),M \in \odot(AKB)$ and $MN,MP,NP$ passes through $A,B,C$ respectively, is a circle centered at the centroid $L$ of triangle $\triangle{A'B'C'}$ passing through $K$ where $A',B',C'$ are the circumcircles of $\odot(BKC),\odot(CKA),\odot(AKB)$ respectively. Another proof of wiseman's generalization: Let $ P', N', M' $ be the midpoint of $ NM, MP, PN $, respectively . Let $ \triangle A_1B_1C_1 $ be the anti-pedal triangle of $ K $ WRT $ \triangle ABC $ . Let $ G_1 $ be the centroid of $ \triangle A_1B_1C_1 $ . Easy to see $ A', B', C' $ is the midpoint of $ KA_1, KB_1, KC_1 $, respectively . Since all $ \triangle PNM $ are similar, so $ PP' $ pass through a fixed point $ P_1 \in \odot (KBC) $ . Similarly, $ NN', MM' $ pass through a fixed point $ N_1 \in \odot (KCA), M_1 \in \odot (KAB) $, respectively . Since $ PP', NN', MM' $ are concurrent at $ G $, so we get $ G, K, P_1, N_1, M_1 $ are concyclic . ie. the locus of $ G $ is a circle $ \mathcal{C} $ passing through $ K $ Consider the case $ P \equiv A_1, N \equiv B_1, M \equiv C_1 $ . Since $ KA_1 $ is the diameter of $ \odot (KBC) $ , so we get $ P_1 $ is the projection of $ K $ on $ A_1G_1 $ . Similarly, $ N_1, M_1 $ is the projection of $ K $ on $ B_1G_1, C_1G_1 $, respectively , so we get $ KG_1 $ is the diameter of $ \mathcal{C} $ . ie. $ L $ is the center of $ \mathcal{C} $ Q.E.D

Even more general, all points $U$ that verify $\triangle MNP \cup U \sim \triangle A'B'C' \cup U'$ orbit on a circle with center $U'$ that passes through $K.$ My proof is exactly the same what I did in my 1st post. If $MU$ cuts $PN$ at $D$ and cuts $\odot(KBC)$ again at $M',$ we have $\angle KM'M=\angle KCM=\angle KAN$ $\Longrightarrow$ $M' \in \odot(KAD).$ Powers of $D$ WRT $\odot(KCA),\odot(KAB)$ are in constant ratio $\overline{DN}:\overline{DP}$ $\Longrightarrow$ center of $\odot(KAD)$ is the intersection $D' \equiv A'U' \cap B'C'$ that verifies $\overline{D'B'}:\overline{D'C'}=\overline{DN}:\overline{DP}.$ Now, powers of $U$ WRT $\odot(KBC)$ and $\odot(KAD)$ are in constant ratio $\overline{UM}:\overline{UD}$ $\Longrightarrow$ locus of $U$ is a circle centered at $U' \in A'D',$ because $\overline{U'A'}:\overline{U'D'}=\overline{UM}:\overline{UD},$ and passing through $K,M'.$ P.S. Telv's method also works in this general configuration.

Here is my solution for this problem Solution Let $F_1$ be first Fermat point of $\triangle$ $ABC$; $I$ be centroid of $\triangle$ $A'B'C'$; $M$ $\equiv$ $IA'$ $\cap$ ($BCA'$), $N$ $\equiv$ $IB'$ $\cap$ ($CAB'$), $P$ $\equiv$ $IC'$ $\cap$ ($ABC'$) ($M$ $\ne$ $A'$, $N$ $\ne$ $B'$, $P$ $\ne$ $C'$) Then: $M$, $N$, $P$ are midpoint of $\stackrel\frown{BC}$, $\stackrel\frown{CA}$, $\stackrel\frown{AB}$ of ($BCA'$), ($CAB'$), ($ABC'$) in $\triangle$ $ABC$ and it's easy to see that $M$, $N$, $P$ are fixed We have: ($F_1M$ ; $F_1P$) $\equiv$ ($F_1M$ ; $F_1B$) + ($F_1B$ ; $F_1P$) $\equiv$ $\dfrac{5 \pi}{6}$ + $\dfrac{5 \pi}{6}$ $\equiv$ $\dfrac{\pi}{3}$ $\equiv$ ($IM$ ; $IP$) (mod $\pi$) So: $F_1$, $I$, $M$, $P$ lie on a circle Similarly: $F_1$, $N$, $I$, $P$ lie on a circle Hence: $F_1$, $N$, $I$, $M$, $P$ lie on a circle or $E$ $\in$ ($INMP$) which is fixed circle

One of the nice problems involving geometric constructions...!

The locus is also known as the Inner Napoleon Circle “History is a set of lies agreed upon.” ― Napoleon Bonaparte

Here is the proof: Construction of the Locus: Let $P$ be the Torricelli-Point of $\triangle ABC$. Then, let $K, L, M$ denote the midpoints of arcs $BC, CA, AB$ of circles $(BPC), (CPA), (APB)$ respectively. Then, we claim that the locus $\L$ of all the possible centroids of $\triangle A'B'C'$ is all the points on $(KLM)$ except point $P$. Claim: The centroid of every such $\triangle A'B'C'$ lies on $\L$. Proof: Let $G$ denote the centroid of $\triangle A'B'C'$. Then, notice that: $A'K, B'L, C'M$ meet at $G$. Hence, $\measuredangle A'GF = \measuredangle KGL = 60^\circ$. Claim: For every point on $\L$, there is an equilateral triangle $\triangle A'B'C'$. Proof: Let $J$ be a point on $\L$. Then, extend lines $JK, JL, JM$ and let it intersect $(BPC), (CPA), (APB)$ at $A', B', C'$ respectively. We will prove that $A', C, B'$ are collinear, from which it follows that $\triangle A'B'C'$ is equilateral by symmetry. Notice that, $\measuredangle GA'C = \measuredangle GB'A = 30^\circ$ and $\measuredangle A'GB' = 120^\circ$. Hence, we must have $\measuredangle A'CB' = 180^\circ$ which implies $A', C, B$ are collinear. Therefore, $\triangle A'B'C'$ is equilateral. Hence proved.