Very nice problem! Let incircle triangle $ABC$ touch sides $BC, AC, AB$ at points $D, E, F$; $I$ is incenter triangle $ABC$ and $I_a, I_b, I_c$ are midpoints of $AI, BI, CI$. From Iran lemma: $A_b, B', C'$ lies on one line; $A_b, D, E$ lies on one line. Similary for points $A_c, B_a, B_c, C_a, C_b$. Claime 1 $B_c, C_b, I_b, I_c, A', D$ are concyclic. Proof: Note that it is Euler circle of triangle $IBC$. $\square$ Similary, $A_c, C_a, I_a, I_c, B', E$ are concyclic; $A_b, B_a, I_a, I_b, C', F$ are concyclic. Claime 2 $(A'B_cC_b), (B'A_cC_a), (C'A_bB_a), (DEF)$ have common point. Proof: After inversion relative incircle we should proof next statement: Let $O$ is circumcentre of triangle $ABC$; $O_a, O_b, O_c$ be reflecting $O$ relative $BC, AC, AB$. Then $(AO_bO_c), (BO_aO_c), (CO_aO_b), (ABC)$ have common point. Generalisation: $O$ can be any point in plane. Let $H$ be orthocenter triangle $ABC$; $B_1, C_1$ are intersection $BH, CH$ with $(ABC)$; $X$ be anti-Steiner $PH$. Obviously, $X, C_1, P_c$ lie on one line; $X, B_1, P_b$ lie one line. Then easy to note that $A, P_b, P_c, X$ are concircle(because $\angle P_cAP_b = 2\angle A$) and $X$ is desired point. Remark: The same result could have been obtained by simply counting the angles, without using the understanding that X is the anti-Steiner of the line $OI$ in original problem(a.k.a Feuerbach point). $\square$ Claime 3 $A_0I \perp B_0C_0$ Proof: Easy see that $\angle I_aXE = \angle I_aB'A = \angle ICA = 90^{\circ} - \angle EDC = 90^{\circ} - \angle DXE \Rightarrow XI_a \perp XD$. Note that $XD$ is rad. axe of $(A'B_cC_b)$ and $(DEF) \Rightarrow DX \perp A_0I; XI_a$ is rad. axe of $(B'A_cC_a)$ and $(C'A_bB_a) \Rightarrow XI_a \perp B_0C_0$. This means that $A_0I \perp B_0C_0$. $\square$ Similary $B_0I \perp A_0C_0$; $C_0I \perp A_0B_0$. Finaly, $I$ is orthocenter triangle $A_0B_0C_0$! $\square$

Let $I$ be incenter of $\triangle ABC; D, E, F$ be projections of $I$ on $BC, CA, AB$. From the hypothesis, we can see that $A_0, B_0, C_0$ are NPC centers of $\triangle IBC, \triangle ICA, \triangle IAB$. Denote $(A_0) \equiv (A'B_cC_b), (B_0) \equiv (B'A_cC_a),$ $(C_0) \equiv (C'B_aA_b)$. Then $(A_0), (B_0), (C_0)$ pass through Feuerbach point $F_e$ of $\triangle ABC$. Suppose that $N, P$ are midpoints of $IB, IC$. Therefore $$(IB_0, IC_0) \equiv (F_eE, F_eF) \equiv (DE, DF) \equiv (IC, IB) \equiv (A'N, A'P) \equiv (F_eN, F_eP) \equiv (A_0B_0, A_0C_0) \pmod \pi$$Similarly, we have $(IC_0, IA_0) \equiv (B_0C_0, B_0A_0) \pmod \pi, (IA_0, IB_0) \equiv (C_0A_0, C_0B_0) \pmod \pi$. Hence $I$ is orthocenter of $\triangle A_0B_0C_0$

Cute! By the problem condition, we are given $A_0,B_0,C_0$ the nine point centers of $\triangle IBC,IAC,IAB$, let $DEF$ be the contact triangle, then as it is well known that feurbach point is the poncelet point of $I$ ,we have $\measuredangle C_0IB_0 = \measuredangle EF_eF = \measuredangle EDF = \measuredangle NF_eP = \measuredangle B_0A_0C_0$, which with cyclic variants is enough to imply that $I$ is indeed the orthocenter of $A_0B_0C_0$

Oops, my knowledge wasn't top shape enough to notice how I could cut down a lot on the proof of these claims. Still, cute problem really enjoyed solving it! Let $D$ , $E$ and $F$ denote the intouch points of $\triangle ABC$ and let $I$ denote the incenter. Further, let $M_a$ , $M_b$ and $M_c$ denote the midpoints of segments $AI$ , $BI$ and $CI$. We start off by revising our previous knowledge. Note that since $C_b$ is the foot of the $C-$altitude of $\triangle BIC$, it is well known that $C_b$ also lies on lines $\overline{EF}$ and $\overline{A'C'}$ by Iran Lemma. Similarly, $B_c$ lies on $\overline{EF}$ and $\overline{A'B'}$, $A_c$ lies on $\overline{DF}$ and $\overline{B'C'}$ , $C_a$ lies on $\overline{DF}$ and $\overline{A'B'}$, $A_b$ lies on $\overline{DE}$ and $\overline{B'C'}$ and $B_a$ lies on lines $\overline{DE}$ and $\overline{A'C'}$. Let $\omega$ denote the incircle of $\triangle ABC$ and $\omega_9$ the nine-point circle. We are now in a position to prove our first claims. Claim : Points $A'$ , $B_c$ , $C_b$ , $M_b$ , $M_c$ and $D$ lie on the same circle $\omega_a$ (and similarly for the other two sides). Proof : It is quite clearly due to the right angles that $BCC_bB_c$ is cyclic with center $A'$. Now, by the midpoint theorem $A'C'\parallel AC$ and $M_bA' \parallel CI$. Since we know that $B_c$ lies on $\overline{A'C'}$, this means that \[\measuredangle M_bA'B_c = \measuredangle ICA = \measuredangle BCI = \measuredangle BCB_c = \measuredangle BC_bB_c = \measuredangle M_bC_bB_c\]which implies that $M_b$ lies on $(A'B_cC_b)$. A similar argument shows that $M_c$ also lies on $(A'B_cC_b)$. Next, note that due to the right angles, $\triangle BID \sim \triangle BCC_b$. Thus, \[\frac{BD}{BC_b}=\frac{BI}{BC}\]from which we can then note that, \[BD \cdot BA' = \frac{BD \cdot BC}{2}= \frac{BI \cdot BC_b}{2}= BM_b \cdot BC_b\]which implies that $D$ also lies on this circle, finishing the proof of the claim. Also, the following startling claim turns out to be true. Claim : Circles $(A'B_cC_b)$ , $(B'A_cC_a)$ and $(C'A_bB_a)$ concur, at the Feuerbach Point $Fe$ of $\triangle ABC$. Proof : We show that $Fe$ lies on $(A'B_cC_b)$ - the others are entirely similar. Let $K = \overline{AI} \cap \overline{BC}$. Since $Fe$ is the tangency of the nine-point circle $\omega_9$ and the incircle $\omega$, it is well known that under inversion centered at $A'$ with radius $A'D$, $Fe$ goes to the second tangency point from $K$ to the incircle. That is, the second intersection of the line through $D$ perpendicular to $AI$ (say $R$). Thus, $Fe$ , $R$ and $A'$ are collinear. Now, \[\measuredangle A'FeD = \measuredangle RFeD = \measuredangle RDA' = \frac{\pi}{2}+ \measuredangle AKB\]Further, \begin{align*} \measuredangle A'B_cD &= \measuredangle CB_cD + \measuredangle A'B_cC\\ &= \measuredangle M_cA'C + \measuredangle B_ACA'\\ &= \measuredangle IBC + \measuredangle ICB \\ &= \measuredangle IBC + \frac{\pi}{2} + \measuredangle IBC + \measuredangle IAB\\ &= \frac{\pi}{2} + \measuredangle KAB + \measuredangle ABK\\ &= \frac{\pi}{2} + \measuredangle AKB \end{align*}and thus, $\measuredangle A'FeD = \measuredangle A'B_cD$ implying that $Fe$ indeed lies on $(A'B_cC_b)$ as claimed. Now, note that, \[\measuredangle M_cFeF = \measuredangle M_cFeD + \measuredangle DFeF = \measuredangle M_CA'C + \measuredangle DEF = \measuredangle IBC + \measuredangle BDF = \frac{\pi}{2}\]What remains now is a radical axis argument. Note that $\overline{FeF}$ is the radical axis of circles $\omega_c$ and $\omega$. Thus, $FeF \perp C_0I$. Further, we observed above that $FeF \perp FeM_c$ so it follows that $C_0I \parallel FeM_c$. Now, we can further note that $\overline{FeM_c}$ is also the radical axis of circles $\omega_a$ and $\omega_b$ as a result of our previous cyclicity claims. Thus, $FeM_c \perp A_0B_0$. Now, since $FeM_c \parallel C_0I$, it finally follows that $C_0I \perp A_0B_0$. A similar argument shows that $B_0I \perp A_0C_0$ and $A_0I \perp B_0C_0$ which indeed implies that the orthocenter of $\triangle A_0B_0C_0$ is simply the incenter $I$ of $\triangle ABC$, as desired.