Let $H$ be the orthocenter of $\triangle ABC$ and let $X$ be the midpoint of $AH$. Then we know that $AXON$ is a parallelogram.
Now, observe that $\angle AOH = \angle AOG$. Now,
$$NO = NA \iff XA = XO \iff \angle AOH = 90^\circ$$
Thus, we are done. ☐

Let $H$ be the orthocenter of $\triangle ABC$ and let $A'$ be the antipode of $A$ in $(ABC)$. Then we know that $ANA'M$ is a parallelogram as the common midpoint of $AA'$ and $NM$ is $O$. We also know that $MH = MA'$.
Now, observe that $\angle AOH = \angle AOG$. Now,
$$NO = NA \iff MA' = MO \iff MA' = MO = MH \iff \angle A'OH = 90^\circ \iff \angle AOH = 90^\circ$$
Thus, we are done. ☐

Say $\angle AOG = 90^\circ$. Let the perpendicular from $N$ to $AO$, meet $AO$ and $AG$ at $D$ and $E$ respectively. Note that $ND\parallel OG$. But we have $O$ as the midpoint of $MN$. So $G$ is the midpoint of $EM$. But we know the centroid divides the median in a ratio $2 : 1$. Hence $E$ is the midpoint of $AG$. Since $E-D-N$ is parallel to $OG$, we get that $D$ is the midpoint of $AO$ and hence $NO = AN$.

For the other direction, say $AN = NO$, define $E$ as the midpoint of $AG$ and $D$ as the midpoint of $AO$. So
$$AE = EG = GM$$
Then note that $DE\parallel OG$ and $NE\parallel OG$ and hence we get $N - D - E$ collinear and hence
$$90^\circ = \angle ADE = \angle AOG$$