Let $T$ be the triangle in the coordinate plane with vertices $\left(0,0\right)$, $\left(4,0\right)$, and $\left(0,3\right)$. Consider the following five isometries of the plane: rotations of $90^{\circ}$, $180^{\circ}$, and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$-axis, and reflection across the $y$-axis. How many of the $3125$ sequences of five of these transformations (not necessarily distinct) will return $T$ to its original position?

As in the problem, $T$ is scalene, so the composition must be the identity. Let $r$ be the $90^\circ$ rotation and write the moves as $r^k$ ($k\in\{1,2,3\}$) or $r^eH$ ($e\in\{0,2\}$, since $V=r^2H$). A composition is a rotation only if it has an even number of reflections, so the five moves have $m=5,3$, or $1$ rotations.

1. \emph{Reduction.} Using $Hr^k=r^{-k}H$, move every $H$ to the right. Each $H$ changes the sign of the exponents that it passes, and the even number of $H$'s cancel at the end. So the composition is $r^N$, where $N$ is a sum of the exponents $k$ and $e$, each with sign $\pm$, and the signs depend only on the positions of the reflections. Since $-e\equiv e\pmod4$ for $e\in\{0,2\}$ and $k\mapsto-k$ permutes $\{1,2,3\}$ mod $4$, the number of choices with $N\equiv0\pmod4$ is the same as the number of choices with $\sum k+\sum e\equiv0\pmod4$.
2. \emph{$m=5$ (no reflections).} Write $k_i=1+d_i$ with $d_i\in\{0,1,2\}$; we need $\sum k_i\in\{8,12\}$, i.e. $\sum d_i=3$ or $7$. The number of solutions with $\sum d_i=3$ is $\binom74-5=30$ (subtract the $5$ solutions having some $d_i=3$), and $d_i\mapsto2-d_i$ shows that $\sum d_i=7$ also has $30$. Total $60$.
3. \emph{$m=3$.} Choose the rotation positions in $\binom53=10$ ways. The two $e$'s have sum $\equiv0$ in $2$ ways and $\equiv2$ in $2$ ways. Triples from $\{1,2,3\}$ with sum $\equiv0$ (sum $4$ or $8$) number $3+3=6$, and with sum $\equiv2$ (sum $6$) number $1+6=7$. So there are $10(2\cdot6+2\cdot7)=260$ sequences.
4. \emph{$m=1$.} Choose the rotation position in $5$ ways. The four $e$'s have sum $\equiv2$ in $8$ of the $16$ cases, and then $k=2$; a sum $\equiv0$ would need $k\equiv0$, which is impossible. This gives $5\cdot8=40$ sequences.
5. The total is $60+260+40=\boxed{360}$. (The same method gives $12$ for three moves, matching the problem.)