Frogs spinning in a pot

In the linear case we get Fibonacci numbers because each of the $n$ frogs they either stay put or exchange position with one of their neighbors. One can think of this as tiling some $1\times n$ space with $1\times 1$ dominos (corresponding to the frogs that stay put) and $1\times 2$ dominos (corresponding to the pair of frogs that exchange positions). Denote $g_n$ to be this number of linear arrangements for $n$ frogs. Then $g_1 = 1$ and $g_2 = 2$, and for each $n\ge 3$, we have $g_n = g_{n-1} + g_{n-2}$, because if the first frog stays put, then we have $g_{n-1}$ many ways to arrange the rest, and if the first frog exchanges with the second frog, then we have $g_{n-2}$ many ways. Hence the recurrence, which is just the Fibonacci numbers (shifted). Here we have $g_{13} = 377$, $g_{14} = 610$.

For the circular case, there are several interpretations of what you may want.

In the simplest situation where the pots are all fixed in place (so we will ignore circular symmetry for now), and assuming the frogs will either stay in place or exchange with one of the neighbors. Denote $a_n$ to be the number of outcomes for $n$ many frogs. Then looking at frog number 1, it can either

(1) stay in place, so the rest of the $n-1$ frogs are in a line, giving $g_{n-1}$ many ways;

(2) exchange with frog 2, giving $g_{n-2}$ many ways;

(3) exchange with frog $n$, giving $g_{n-2}$ many ways.

So this gives $a_n = g_{n-1} + 2g_{n-2}$.

Now, there is also the situation where all the frogs jump to the left, or all the frogs jump to the right. Then in this case we have $a_n = g_{n-1} + 2g_{n-2} +2 $. (However if you take rotational symmetry into consideration, these two extra configurations are just rotationally the same no frog moved.) With $n = 15$, it seems to agree with what you got.

However, as the circular dish is spinning, it seems to me that we should take rotational symmetry into consideration, then we will get $a_n = g_{n-1} + 2g_{n-2}$.

If there are any other missing ways of how the frogs can move, then we will need to account for those. Otherwise we have our result.