Looking at the cube from "the front", i.e. side $ABFE$,and considering this a square in the plane, let $A | D$ be in the lower left-hand corner and let it have coordinates $(0, 0)$, $B | C$ at $(1,0)$, $F | G$ at $(1,1)$ and $H | E$ at $(0,1)$. Reflect the square around each side until the whole plane is tiled. Let $o$ stand for odd, and $e$ stand for even and a $X|Y$-group is a reflection of $X | Y$. Then we see that $A | D$-group is at $(e, e)$, $B | C$-group at $(o,e)$, $F | G$-group at $(o,o)$ and $H | E$-group at $(e,o)$. Now project the movement of the beam perpendicularly onto this square $ABFE$. Transform this in our tiled plane to obtain a line, where this line starts at $(0,0)$, and each original reflection is the line leaving a unit square of this plane (in other words keeping the line but reflecting the squares). This line can be given the equation $y_1= \frac{a_1}{b_1}\cdot x_1; a,b\in \mathbb{N}$, since this line should eventually reach a lattice point and we thus must have that the parameter of $x_1$ is rational. In fact it reaches lattice point $(x_1, y_1)=(b_1,a_1)$, from which it is clear that we also want $\gcd(a_1,b_1)=1$ (otherwise there would be a lattice point reached before $(b_1, a_1)$ at which the beam would already have left). This point $(b_1, a_1)$ is a $X | Y$-group and corresponds to $(\text{par}(b_1), \text{par}(a_1))$ of our original cube vertices ($\text{par}(x)$ gives the parity of some integer $x$). Now also note that considering $ABFE$ we have exactly $a_1 + b_1 -2$ reflections projected onto it. The above projections, tiling of plane etc. can also be done for $BCGF$ and $HEFG$. For $BCGF$: $A | E $ at $(0,0)$ and $A | E$-group at $(e,e)$, $H | D$ at $(0,1)$ and $H|D$-group at $(e,o)$, $B|F$ at $(1,0)$ and $B|F$-group at $(o,e)$, $G | C$ at $(1,1)$ and $G|C$-group at $(o,o)$. Also $y_2=\frac{a_2}{b_2}x_2$ and $a_2+b_2-2$ reflections. For $HEFG$: $A | B $ at $(0,0)$ and $A | B$-group at $(e,e)$, $E | F$ at $(0,1)$ and $E|F$-group at $(e,o)$, $C|D$ at $(1,0)$ and $C|D$-group at $(o,e)$, $G | H$ at $(1,1)$ and $G|H$-group at $(o,o)$. Also $y_3=\frac{a_3}{b_3}x_3$ and $a_3+b_3-2$ reflections. Since our beam is not parallel to a side of the cube, we must have that the number of projected reflections must be equal for two of the sides. Now consider the following two cases: Case 1: If $n$ is even, then $a_{i_1}, b_{i_1}$ and $a_{i_2}, b_{i_2}$, $i_1, i_2 \in \{1,2,3\}, i_1\neq i_2$, are all odd (since $\gcd(a_{i_1}, b_{i_1})=\gcd(a_{i_2}, b_{i_2})=1$. Then we can check that $G$ is the only possible vertex in that case, which can be achieved by taking $(a_{i_1}, b_{i_1})=(1,n+1)$ and $(a_{i_2}, b_{i_2})=(n+1,1)$, Case 2: If $n$ is odd, then one of $a_{i_1}, b_{i_1}$ is odd and one is even, same for $a_{i_2}, b_{i_2}$. Then we can check that all vertices except $A$ and $G$ work, where a specific vertex can be reached by taking two specific projection sides and a corresponding combination of odd and even $a_{i_1}, b_{i_1}$ and $a_{i_2}, b_{i_2}$, with one of $\{a_{i_j}, b_{i_j}\}$, $j \in \{1,2\}$, equal to $2$ and one equal to $n$. In conclusion, for $n$ even the only solution is $G$ and for $n$ odd the solutions are $\{B,C, D, E, F, H\}$. $\square$