This question is about the understanding of the symmetry of Nash game, based on the definition given in A note on Mean Field Games. The definition is given at the beginning of the note in the static game, and shall be important for the understanding of mean field effect (MFG) in dynamic setting. In this below, I formulated two questions [Q1] and [Q2].
For simplicity, we set
$N = 3$ is the number of players;
$x_i\in \mathbb R$ is the player i's state;
$F_i(x_1, x_2, x_3) \in \mathbb R$ is the player i's cost;
-QED-
The definition of symmetry is given in Section 2.1 as follows.
[Def] The game is called symmetric if $$F_i(x_1, x_2, x_3) = F_{\sigma(i)}(x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}),$$ for any permutation $\sigma$ on $\{1, 2, 3\}$.
-QED -
The following example is a special case of symmetric game discussed in Section 2.2.
[ex] $F_i(x_1, x_2, x_3) = x_i$ for $i = 1,2,3$.
[Q1] It seems to me that [ex] is not symmetric game according to [Def], because if we check symmetry with $\sigma(1, 2, 3) = (2, 3, 1)$, then we have $F_1(x_1, x_2, x_3) = x_1$, while $F_{\sigma(1)}(x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}) = x_3$. Am I missing something?
-QED -
[Q2] If the above argument were correct, I guess it is somehow a typo of $\sigma$ in place of $\sigma^{-1}$, but I am not sure? See the new definition here.
[Def-1] The game is called symmetric if $$F_i(x_1, x_2, x_3) = F_{\sigma^{-1}(i)}(x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}),$$ for any permutation $\sigma$ on $\{1, 2, 3\}$.
-QED -
Under the [Def-1], one can show that the game in [ex] is symmetric, since $$F_{\sigma^{-1}(i)}(x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}) = x\circ \sigma(\sigma^{-1}(i)) = x(i).$$
[Comment to Replies] An example on symmetric game in the same spirit "a player's payoff is equal to his own action, regardless actions of others" is recommended as follows: \begin{equation}\label{eq:1} F_{\sigma(i)} (x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}) = x_{\sigma(i)}, \quad \forall \sigma \quad\quad\quad\quad (eq1). \end{equation} However, this representation is still confusing. Let's try to evaluate $F_1(1, 2, 3)$ according to (eq1).
First, if we take the identity permutation $\sigma (1, 2, 3) = (1, 2, 3)$ and $i=1$ for (eq1), we have $F_1(x_1, x_2, x_3) = x_1$, which means $F_1(1, 2, 3) = 1$.
Second, if we apply $\sigma(1,2,3) = (2, 3,1)$ and $i=3$, then we have $F_1(x_2, x_3, x_1) = x_1$, which means $F_1(1, 2, 3) = 3$.
In other words, $F_1(1, 2, 3)$ is not well defined according to (eq1)?
The definition is correct, but the example is not properly stated (leaving the reader to fill the details).
Presumably, the example states that each player $i$ has a payoff equal to his own action $x_i$, regardless of what actions other players are choosing. Formally speaking, this should be stated as $$F_{\sigma(1)}(x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)}) = x_{\sigma(1)}$$ for each $i=1, 2, \ldots, n$. This definition is consistent with the general definition of symmetry and eliminates all pending issues in your OP.