= Solution
In a <finite game>, a profile of independent <mixed strategies> $\sigma=(\sigma_1,\ldots,\sigma_n)$ is a <Nash equilibrium> if no player can improve their expected payoff by changing only their own <mixed strategy>. In symbols,
$$
U_a(\sigma_a,\sigma_{-a})\geq U_a(\tau_a,\sigma_{-a})
\quad\text{for every player }a\text{ and every strategy }\tau_a.
$$
Because expectation is linear in a player's own <mixed strategy>, checking this inequality for all their <pure strategies> suffices. Nothing requires the players' payoffs to sum to zero.
For the <symmetric finite game>, write $e_i(p)=e(i,p)$ and $\bar e(p)=\sum_i p_i e_i(p)$. Each $e_i(p)$ is a polynomial in the opponents' independent probabilities and therefore continuous on the <probability simplex> $\Delta=\{p\geq0:\sum_i p_i=1\}$. Put
$$
r_i(p)=\max(0,e_i(p)-\bar e(p)),\qquad
S(p)=\sum_i r_i(p),\qquad
F_i(p)=\frac{p_i+r_i(p)}{1+S(p)}.
$$
This continuous map has nonnegative coordinates summing to one, so it maps the compact <convex set> $\Delta$ into itself. The <Brouwer fixed-point theorem> gives $p=F(p)$. At that fixed point $r_i=p_iS$. If $S>0$, every $i$ in the <strategy support> has $r_i>0$ and therefore $e_i(p)>\bar e(p)$. But
$$
\sum_i p_i(e_i(p)-\bar e(p))=0,
$$
whereas every supported term would be strictly positive, a contradiction. Thus $S=0$, all pure payoffs are at most $\bar e(p)$, and their weighted average equals $\bar e(p)$. Every supported <pure strategy> is consequently a <best response> to the common opposing <mixed strategy>. \b[The profile in which every player uses $p$ is a symmetric Nash equilibrium.] This proves the <gain-map proof of symmetric equilibrium> for any finite number of players, including boundary fixed points of the <probability simplex>.
In the <least unique bid auction>, let $p$ be the common <probability> of bid 1. Bidding 1 wins only when the other two bids are both 2, and its winning payoff is $3-1=2$. Bidding 2 wins only when the other two bids are both 1, and its winning payoff is $3-2=1$. Hence the two expected payoffs are
$$
e_1(p)=2(1-p)^2,\qquad e_2(p)=p^2.
$$
Neither symmetric <pure strategy> profile is an equilibrium: against two bids of 1 a deviation to 2 earns 1 instead of 0, and against two bids of 2 a deviation to 1 earns 2 instead of 0. An interior <symmetric equilibrium> must make the two expected payoffs equal. Solving $2(1-p)^2=p^2$ leaves just one root in $(0,1)$, giving
$$
\boxed{p=2-\sqrt2\text{ for bid 1},\qquad1-p=\sqrt2-1\text{ for bid 2}.}
$$
The expected payoff to each player is $p^2=6-4\sqrt2$.
To count all nonsymmetric <Nash equilibria>, let $p_i$ be player $i$'s <probability> of bidding 1. The difference between their expected payoffs from bids 1 and 2 is
$$
\Delta_i=2(1-p_j)(1-p_k)-p_jp_k
=2-2p_j-2p_k+p_jp_k.
$$
The <best response> conditions are $p_i=1$ when $\Delta_i>0$, $p_i=0$ when $\Delta_i<0$, and arbitrary $p_i$ when $\Delta_i=0$.
For any $0\leq t\leq1$, the profile $(p_1,p_2,p_3)=(1,0,t)$ satisfies these conditions: $\Delta_1=2(1-t)\geq0$, $\Delta_2=-t\leq0$, and $\Delta_3=0$. Thus every permutation of this profile is a <Nash equilibrium>. At $t=0,1$ these give the six pure profiles with at least one bid of each kind; for $0<t<1$, one player mixes freely and the other two use opposite <pure strategies>.
There are no other nonsymmetric equilibria. If all three probabilities are interior, every $\Delta_i$ must vanish. Subtracting two such equations gives
$$
\Delta_1-\Delta_2=(p_1-p_2)(2-p_3)=0.
$$
Since $2-p_3>0$, all probabilities are equal, yielding the symmetric solution already found. If exactly two players mix and the third uses bid 2, either mixer strictly prefers bid 1 because $\Delta=2(1-p_j)>0$. If the third uses bid 1, either mixer strictly prefers bid 2 because $\Delta=-p_j<0$. Both cases are impossible. If exactly one player mixes, their indifference requires the other players to use opposite bids. Finally, with no mixing, the all-equal pure profiles fail and the remaining six satisfy the <best response> conditions.
Consequently the complete <Nash equilibrium> set consists of the single symmetric mixed profile together with the six closed one-parameter families given by permutations of $(1,0,t)$. Their endpoints overlap at the six pure equilibria. In particular,
$$
\boxed{\text{there are uncountably infinitely many nonsymmetric equilibria}.}
$$
Counting only the six pure ones would miss the interior points of these families in the <two-bid three-player least unique bid auction>.
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