What is 2nd round of rationalizability for a 1st price auction? Consider a first-price auction.
Suppose that we have $N$ bidders, and they believe that their opponents' values is drawn from a uniform distribution on interval $[0,1]$.
Let us eliminate weakly dominated strategies.
The 1st round will clearly eliminate all bids higher than the private value $x$. But what range of prices will be eliminated in a 2nd round?
My conjecture: after elimination of bids higher than private values, bidder $i$'s objective function in a 2-bidder situation will be $(v_i-b_i)\Pr(b_{-i}\leq b_i)$. The probability $\Pr(b_{-i}\leq b_i)$ is maximized when $b_{-i}$ is approaching $v_{-i}$. So maximized form of objective functions of bidder $i$ is $(v_i-b_i)\Pr(v_{-i}\leq b_i)$, which is $(v_i-b_i)b_{i}$. (Since we assume uniform distribution on values) So after 1st round of rationalizability, a bidder's maximized payoff will be $\frac{v^2_i}{4}$. This means that in 2nd round of rationalizability, any bidder will not be bidding higher than $v_i-\frac{v^2_i}{4}$.
Elimination of (weakly) dominated actions will not get you far. In fact, all bids strictly between 0 and the value are undominated. Hence, a buyer's optimal bid in a first-price auction can't be determined without knowing the bidder's belief (unless the buyer's value is 0).
Claim: For any type $x&gt;0$, all bids $c \in (0,x)$ are undominated.

$c$ is not dominated by any bid $b&gt;c$: $b$ yields a strictly smaller
than $c$ if all other bidders bid $b_j&lt;c$.
$c$ is not dominated by any bid $b&lt;c$: $b$ yields a strictly smaller utility than $c$ if all other bidders bid $b_j=c$.

You can show that, for all types $x&gt;0$, bidding zero is dominated by any $b\in (0,\frac{N-1}{N} x)$, though.
Suppose $p$ is the probability that all other $(N-1)$ bidders bid zero. Then, bidding zero yields utility $\frac{p}{N}x$: winning by a randomization because everyone bids the same and then winning the lottery and paying nothing.
However, when bidding $b$ from the interval above, your exepected payoff is at least
$$\geq p(x-b) &gt; p (x - \frac{N-1}{N} x) &gt; \frac{p}{N}x.$$