The proof is based on a well-known fact recalled in the lemma below.

Lemma. Let $ABC$ be a triangle and let $O$ and $\omega$ be its circumcentre and nine-point centre, respectively. Then the reflexion $O'$ of $O$ in the line $BC$ lies on the line $\omega A$, and $\omega$ is the midpoint of the segment $O'A$.

If $\Gamma$ and $\Gamma'$ do not coincide, then they are clearly the reflexion of one another in the line $YZ$, and, consequently, so are their centres. Let $O'$ be the centre of $\Gamma'$. By the lemma, the nine-point centre of the triangle $XYZ$ is the midpoint of the segment $O'X$ which in turn is the centre of $\gamma$ and the conclusion follows.

If $\Gamma$ and $\Gamma'$ coincide, recall that the homotheties mapping $\gamma$ to $\Gamma$ are: one of ratio $2$ centred at the orthocentre $H$ of the triangle $ABC$; and one of ratio $-2$ centred at the centroid of the triangle $ABC$. Consequently, $X = H$, so the triangle $ABC$ is right-angled which case is ruled out by hypothesis.