Some simple notations
Let $I = \{1, \ldots, N\}$ and consider a $N$ player game with strategies $S = \{1, \ldots, n \}$.
Let $e^j_i = [0, \ldots, \underline{1}_\text{jth position}, \ldots 0]$ denote $jth$ pure strategy of player $i$
Let $f$ denote the payoff function
Finally let $\text{supp}(x_i) = \{ j \in S| x_i^j > 0\}, x_i$ is mixed strategy of $ith$ player
Then, claim. $x^* = (x_1^*, x_2^* \ldots, x_N^*)$ is a mixed symmetric Nash equilibria if:
$$f_i(e^j_i,x_i^*) = f_i(x_i^*,x_i^*), \forall j \in \text{supp}(x^*), \forall i \in I$$
The application of this shows up in showing that any NE is a stable point of the replicator dynamics
Assume a single population. Consider the replicator dynamics $$\dot\mu_i(t) = \mu_i(t)(l_i(\mu) - \mu^Tl(\mu))$$
where $l_i(\mu) = f(e^i,\mu), \mu^Tl(\mu) = f(\mu,\mu)$.
Then if $\mu$ is a symmetric Nash equilibria, $l_i(\mu) = \mu^Tl(\mu) \implies \dot\mu_i(t) = 0$
It doesn't seem to be very intuitive that the pure strategy would yield the exact same payoff as mixed strategy, but running this theorem against some well known games there seems to be no violation.
Can someone show me how the claim is derived and how it applies to the replicator dynamics proof?
By the definition of Nash equilibrium, no player can improve their payoff by playing a different strategy. However, if the strategies of all the other players are viewed as fixed, then the payoff of a mixed strategy is linear (and although we tacitly assume that the weights of the coefficients sum to $1$ so that we can interpret them as the probabilities of playing pure strategies, it is mathematically convenient to remove that restriction and just think about linear maps).
To ease notation, let us assume that player $i$ has $2$ pure strategies, $e_1$ and $e_2$, that we have some Nash equillibrium, and in that state we are fixing all the other player's strategies, so we can look exclusively at player $i$'s payoff function of his own strategy. Let $P_1=P(e_1), P_2=P(e_2)$, and $P=P(x^*)$ be the payoffs of the pure strategies and the equillibrium strategy, resepctively. If $P_1>P_2$, then $P_1>aP_1+bP_2$ for any $a,b$ such that $a+b=1, 0\leq a<1$, so if $e_1$ and $e_2$ are both in the support of $x^*$, we must have that their payoffs are equal (subject to fixing everybody else's strategies at the equilibrium). In this case, every mixed strategy has the same payoff.
One can easily generalize this argument (which is essentially about maximizing a linear function on a simplex) to show that pure strategies can only occur in the support if their payoff (all other things remaining equal) is the same as the mixed strategy.
Note that none of the above makes any mention of the payoff function being symmetric. It also does not require the game to have two players.
The last bit of the puzzle, which I have not yet worked out all the details for, is why we can limit ourselves in a symmetric game to assuming both players are using the same strategy. It is not immediately clear to me why if $(x^*,y^*)$ is a Nash equillibrium then $(x^*,x^*)$ should be as well. However, if you can show this, then the result follows.