We will first prove the following result: Lemma.Let ABC be an acute-angled triangle and H its orthocenter. Let T be a point on the circumcircle of ABC and P the orthocenter of AHT. If D is a point such that ABDC is a parallelogram , E is a point such that ATPE is a parallelogram and N is the intersection point of HP and BC,then the points D,N,E are collinear. Proof:Let M be the midpoint of the segment BC and R the reflection of T with respect of M.Since BTCR is a parallelogram we get that the angles BRC and BTC are equal.But ACBT is cyclic,so BRC and BAC are equal angles.This means that BHCR is cyclic.A simple angle-chasing shows that HR is perpendicular to AT,so P,H,N,R are collinear points. On the other side,TP is parallel to BC,so MN is the midline of TPR,which means that MN is half of BC.Since AE is parallel to BC and equal to TP, it follows that MN is also midline in ADE.Hence D,N,E are collinear. We come back to the problem. It is easy to see that P is the orthocenter of TDO,where T is the reflection of M with respect to the midpoint of the segment DP. Let X be a point on one of the sides of DP (which side depends on the measure of MON) such that it lies on the perpendicular bisector of DP and the distance to DP is half of DM.Let the circle of center X and radius XD intersect the euler circle of ABC for the second time at T' and let P' the orthocenter of DOT'.It's easy to see that DMP'T' is a parallelogram.Also notice that O is the orthocenter of DEF and DEAF is a parallelogram.From the lemma we get that OP' ,AM and EF are concurent,so N lies on OP'.But then it's obvious that P and P' coincide,so T and T' coincide.

Is there any solution by moving point??

Let $P'$ be the reflection of P in $N$,and $X$ be the reflection of M in the midpoint of $DP$ then $AP'MP$ is a parallelogram, since $$90^{\circ}=\angle AP'O=\angle AFO=\angle AEO$$hence $A,P',O,F,E$ are concyclic. Now since $MPDX$ is a parallelogram, $AP'DX$ is a parallelogram, meanwhile $AFDE$ is a parallelogram. Let $R$ be the midpoint of $AD$, then a homothety at $R$ with ratio $-1$ will sent $AP'FE$ to $DXEF$, therefore $X$ lies on the circumcircle of $\triangle DEF$, that is the nine-point circle. In fact this argument also works if $M$ is not on $BC$, then $N$ will be the midpoint of $AM$ and $P$ will be the projection of $M$ on $NO$

Second solution by complex numbers: Set up the complex plane with $(ABC)$ as the unit circle, since $N$ is the midpoint of $AM$, $$n=\frac{a+m}{2}$$Since $P$ is the projection of $M$ on $NO$, $$p=\frac{\overline{n}m+n\overline{m}}{2\overline{n}}=\frac{m}{2}+\frac{n\overline m}{2\overline n}$$Since $PQDM$ is a parallelogram,$p-m,=q-d$, where the reflection of $M$ w.r.t. the midpoint of $DP$ is denoted by $Q$, $$q=d+p-m=\frac{b+c}{2}+\frac{m}{2}+\frac{n\overline m}{2\overline n}-m=\frac{b+c}{2}-\frac{m}{2}+\frac{n\overline m}{2\overline n}$$It is well-known that the nine-point center of $\triangle ABC$ is $$\frac{a+b+c}{2}$$Now we have \begin{align*} \frac{a+b+c}{2}-q&=\frac{a}{2}-\frac{n\overline m}{2\overline n}+\frac{m}{2}\\ &=n-\frac{n\overline m}{2\overline n}\\ &=n(\frac{2\overline n-\overline m}{2\overline n})\\ &=n(\frac{\overline a+\overline m-\overline m}{2\overline n})\\ &=\frac{n\overline a}{2\overline n} \end{align*}Hence $$|\frac{a+b+c}{2}-q|=|\frac{n}{\overline n}||\frac{a}{2}|=\frac{1}{2}$$Therefore we conclude that $Q$ lies on the nine-point circle.