I like this problem Claim 1: $G$ is the A-sharkydevil point. Proof: Let $F'$ the touch point of the incircle of $\triangle ABC$ with $BC$ and let $N'$ the midpoint of the minor arc $BC$ of $\Gamma$ using I-E lemma we do an inversion w.r.t. $(BIC)$ so $N'$ is the center of inversion, then $E$ maps to $E'$ which lies on $BC$ and let $A'$ a point on $\Gamma$ such that $AA' \parallel BC$, now since its clear that $N'D$ passes through $A'$ we have that $\angle E'N'D=90=\angle E'F'D$ hence $E'F'N'D$ is cyclic so inverting back we get that the inverse of $F'$ is the intersection of $ED$ and $\Gamma$ but hey, $G$ is that point and since its well known that the A-sharkydevil point lies on $F'N'$ the claim is proven. Finishing: Let $NF \cap \Gamma=X$ then we make an inversion w.r.t. $(BIC)$ and reflecting over $NN'$ so $N$ is sent to $M$, $F$ is sent to $G$ which means that $X$ is sent to $(GMN') \cap BC=X' \ne M$ now since $\angle N'MF=90=\angle N'XF$ we have $N'MFX$ cyclic so inverting back we get $G,N,X'$ colinear and by angle chasing $$\angle NX'N'=\angle N'MX=\angle N'FX \implies N'X'NF \; \text{cyclic}$$And inverting back gives $G,M,X$ colinear thus we are done

Indeed a very beautiful problem! Firstly, let $K=ID \cap BC$ (The point where the incircle of $\triangle ABC$ touches $BC$), $L=MN\cap \Gamma$ (The midpoint of arc $BC$ not containing $A$). Claim: $L,K,G$ are collinear Proof: If $G'=KL\cap \Gamma$ and $E'=GD \cap \Gamma$, we need that $G'=G \iff E'=E \iff AE' \perp BC \iff AE' \parallel ID$. Now, notice that $BC$ is the radical axis of $\Gamma$ and $(BIC)$, so: $$Pot_{\Gamma}K=Pot_{(BIC)}K \iff KG' \cdot KL=KI \cdot KD \iff ILDG' \textit{ is cyclic} $$ So, in $\Gamma$: $\angle ILG=\angle AG'E'$ In $(G'ILD)$: $\angle ILG'=\angle IDG' \iff \angle IDG' =\angle AE'G' \iff AE' \parallel ID$ $\square$ Finally, sine $G,K,L$ are collinear, $L,M,N$ are collinear and $M$ is midpoint of both $BC$ and $KF$, we are done by butterfly theorem. $\square$

Let $K$ be the midpoint of the other arc $BC$ in $\Gamma$ ,$T=ID \cap BC$, the projection of $I$ in $BC$, and $H=GM \cap FN$. We need to show that $H$ lies on $\Gamma$. From the cyclic quadrilaterals and because $AE \parallel ITD$ we get that: $\angle DGK = \angle EGK = \angle EAK = \angle EAI = \angle DIK$, hence $DGIK$ is cyclic. By radical axis on $(DGIK), (GBKC), (DBIC)$, we get that $GK, BC, ID$ are concurrent and so $G,T,K$ are collinear, from where we conclude that $\angle NGT=\angle NGK = 90= \angle NMT$ and hence $NGTM$ is cyclic. It is well known that $F,T$ are symmetric with respect to $MN$ and so using the symmetry and the cyclic quadrilaterals: $\angle KNH = \angle MNF = \angle MNT = \angle MGT = HGK$, and $H$ lies on $\Gamma$, as needed.

Cheesed with ratios Claim. $G$ is the $A$-Sharkydevil point. Proof. Let $I'$ be on $(BIC)$ such that $II' \parallel BC$, and let $EI'$ meet $\Gamma$ again at $G'$. Also let $S$ be midpoint of arc $BC$ not containing $A$. It is easy to angle chase to get $EN \parallel SI'$, and $S$ is the midpoint of $DI'$, so \[ (G,G';N,S) \stackrel E= (D,I'; \infty_{DI'}, S) = -1 \]implies $G'$ and $G$ are reflections across $NS$. Considering $EI'G'$, this implies $GI$ passes through the $A$-antipode, so $G$ is the Sharkydevil point. $\square$ Now it suffices to show that $(B,C;G,N)=(B,C;M,F)$ i.e. $BG/GC=CF/FB$. This is equivalent to the well-known fact that the $A$-intouch point lies on the angle bisector of $\angle BGC$, hence proved.

Another interesting Sharkydevil property! We start off (as in all the above solutions) by characterizing $G$. First, let $A'$ be the $A$-antipodal point and $M_A$ be the minor arc midpoint of $BC$. Claim : Point $G$ is the $A-$Sharkydevil point of $\triangle ABC$. Proof : We first note the following congruency. Since it is well known that $M_A$ is the center of $(BIC)$, $M_AD=M_AI$. Further, it is well known that $EA' \parallel BC$ so $M_A$ is also the minor arc midpoint of $EA'$. Thus, $M_AE=M_AA'$. Now, note that since $A-I-M_A$ by definition, it follows that $\measuredangle IM_AA' = 90^\circ$. Then, \begin{align*} \measuredangle DM_AE &= \measuredangle DM_AB + \measuredangle BM_AE\\ &= 2\measuredangle DIB + \measuredangle BAE \\ &= 2(90+\measuredangle CBI) + 90 + \measuredangle ABC\\ &= 2\measuredangle CBI + 90 + \measuredangle ABC\\ &= \measuredangle CBA + \measuredangle ABC + 90\\ &= 90^\circ \end{align*}Thus, $\measuredangle DM_AE = 90^\circ = \measuredangle IM_AA'$ and it follows that $\triangle DM_AE \cong \triangle IM_AA'$. In particular, $\measuredangle M_AED = \measuredangle M_AA'I$ which we shall use extensively in the following angle chase. We let $G' = \overline{DE} \cap \overline{IA'}$. Then, \begin{align*} \measuredangle A'G'E &= \measuredangle A'ED + \measuredangle GA'E\\ &= \measuredangle A'ED + \measuredangle IA'E\\ &= \measuredangle A'EM_A + \measuredangle M_AED + \measuredangle IA'M_A + \measuredangle M_AA'E\\ &= \measuredangle A'EM + \measuredangle M_AA'E\\ &= 2\measuredangle A'EM_A\\ &= 2\measuredangle A'AM_A\\ &= \measuredangle A'AE \end{align*}which which it clearly follows that $G'$ lies on $(ABC)$. Thus, $G$ is the intersection of line $\overline{IA'}$ and $(ABC)$ which is well known to be the $A-$Sharkydevil point as claimed. Now, let $P$ be the foot of the altitude from $I$ to $BC$ ($A-$incircle tangency point). Then, it is well known that $F$ is the reflection of $P$ across $M$ and that points $G$ , $P$ and $M_A$ are collinear. Now, let $X_1 = \overline{NF} \cap (ABC)$. Clearly, $MM_AX_1F$ is cyclic due to the right angles. Then, let $X_2 = \overline{GM} \cap (ABC)$. We now have, \begin{align*} \measuredangle M_AX_2M &= \measuredangle M_AX_2G\\ &= \measuredangle M_ACG \\ &= \measuredangle M_ACB + \measuredangle BCG\\ &= \measuredangle M_AGB + \measuredangle BCG \\ &= \measuredangle CGM_A + \measuredangle BCG\\ &= \measuredangle BPG\\ &= \measuredangle FPM_A\\ &= \measuredangle M_AFM \end{align*}so, $X_2M_AMF$ is also cyclic. From this, it clearly follows that $X_1=X_2$, which implies that lines $\overline{GM}$ and $\overline{NF}$ concur on $(ABC)$ as desired.

Claim 1: $G$ is $A$-sharkydevil point. As always we invert about $(BIC)$ then $(ABC) \leftrightarrow BC$,$ A \leftrightarrow A'$. Now if $T$ is incircle touchpoint with $BC$ then, we need to show $G \leftrightarrow T$. Let $L$ be center of $(BIC)$ and $E'= EL \cap BC$, then $E \leftrightarrow E'$ $G,E,D$ collinear $\Rightarrow$ $G',E',D,L$ cyclic. So we prove $E',T,L,D$ are con-cyclic $$\angle LDT = \angle CDI - \angle CDL = \frac{\angle B}{2} - \frac{\angle C}{2}$$$$\angle LE'C = 180 - \angle CLE - \angle LCB = \angle LDT$$hence we get $T \equiv G'$.. Let $X = NT \cap (ABC)$. It's well known that $(G,X;B,C)=-1$ Now as $T,F$ are symmetric with $NM$, if $K = NF \cap (ABC)$ then $XK \parallel BC$. $$(B,C;M,\infty{BC})\stackrel{K}{=}(B,C;M',X)=-1=(B,C;G,X)$$hence $G,M,K$ collinear.

Let $G'$ be $A$ sharkydevil point. $\Delta RST$ be the intouch triangle. $NR,NF$ meet $(ABC)$ again at $H',H$. $N',A'$ be the antipode of $N,A$.Clearly $G'-I-A'$ are collinear. By spiral sim at $G'$, $\frac{G'B}{G'C}=\frac{BT}{CS}=\frac{BR}{CR}$ ,hence by angle bisector theorem $G'-R-N'$ are collinear. $-1=(N,N';B,C)\stackrel{R}{=}(H',G';C,B) \implies G'H'$ is symmedian in $G'BC$ . Since $H',H$ are symmetrical wrt $NN'$ we obtain $G'-M-H$ collinear. Let $IR$ meet $(AI)$ again at $J$, since $IJ \parallel N'N$ we obtain $G'-J-N$ collinear by reims theorem. Let $I'$ be $A$ excenter , we know $AJ \parallel BC \parallel D'I'$. Thus $(J,R;I,G'E \cap IR)\stackrel{G'}{=}(N,N';A',E)=-1=(A,AI \cap BC;I,I')\stackrel{\infty_{BC}}{=}(J,R;I,D)$. Thus $G'E \cap IR=D$.Thus $G'=G$