How many ways to read the palindromic word ROTATOR in this design?
Here is a slightly easier argument.
Let's just count how many ways there are to go from an R to an A. The R's on one side give this:
$$R^1 \quad R^1 \quad R^1\\ O^1 \quad O^2 \quad O^2 \quad O^1 \\ T^3 \quad T^4 \quad T^3 \\ A^3 \quad A^7 \quad A^7 \quad A^3$$
Starting at the R's on the other end gives this:
$$A^4 \quad A^7 \quad A^7 \quad A^4 \\ T^1 \quad T^3 \quad T^4 \quad T^3 \quad T^1 \\ O^1 \quad O^2 \quad O^2 \quad O^1 \\ R^1 \quad R^1 \quad R^1$$
Adding those together, there are $7$, $14$, $14$, and $7$ ways to go from any R to each of the central A's. Conversely there are the same number of ways to go from each of those A's back to any R. Combining any R-to-A path with any A-to-R path from the same A we get $7\cdot7+14\cdot14+14\cdot14+7\cdot7 = 490$.
As @Peter hinted in a comment, we have to count for paths going from left to right, right to left as well as left to central $A$ column and back to left and so on.
Notation : $^{m}T^{n}$ indicates $T$ was first reached in $n$ number of ways and later $m$ total number of ways. Read right number first, left number second always.
Left to right paths (unidirectional): $$ R \quad R \quad R\\ O^{1} \quad O^{2} \quad O^{2} \quad O^{1} \\ T^{3} \quad T^{4} \quad T^{3} \\ A^{3} \quad A^{7} \quad A^{7} \quad A^{3} \\ T^{3} \quad T^{10} \quad T^{14} \quad T^{10} \quad T^{3} \\ O^{13} \quad O^{24} \quad O^{24} \quad O^{13} \\ R^{37} \quad R^{48} \quad R^{37} $$
Left + Right $= 2(37+48+37)=2\cdot 122 = 244$
Left to center and back to left : (top to bottom to top) $$ ^{34}R^{0} \quad ^{48}R^{0} \quad ^{34}R^{0}\\ ^{10}O^{\color{red}{1}} \quad ^{24}O^{\color{red}{2}} \quad ^{24}O^{\color{red}{2}} \quad ^{10}O^{\color{red}{1}} \\ ^{10}T^{3} \quad ^{14}T^{4} \quad ^{10}T^{3} \\ A^{3} \quad\quad A^{7} \quad\quad A^{7} \quad\quad A^{3} $$
Left to center, back to left : $34+48+34=116$
Right to center and back to right : (bottom to top to bottom) $$ A^{4} \quad\quad A^{7} \quad\quad A^{7} \quad\quad A^{4} \\ ^{4}T^{1} \quad ^{11}T^{3} \quad ^{14}T^{4} \quad ^{11}T^{3} \quad ^{4}T^{1} \\ ^{15}O^{\color{red}{1}} \quad ^{25}O^{\color{red}{2}} \quad ^{25}O^{\color{red}{2}} \quad ^{15}O^{\color{red}{1}} \\ ^{40}R^{0} \quad ^{50}R^{0} \quad ^{40}R^{0} $$
Right to center, back to right : $40+50+40=130$
Total ways : $244+116+130=490$