A Fermat pseudoprime for base $b\geq 2$ is a composite integer $n$ such that $$b^{n-1} - 1\equiv 0\: \left(\mathrm{mod}\:n\right)$$ Looking at the list of Fermat pseudoprimes for bases $b \leq 1024$ here, it appears as if the first pseudoprime for a given base is usually "small." For $b = 2$, the first Fermat pseudoprime is $n = 341$. But the next base $b$ where the first Fermat pseudoprime is $\geq 341$ is $b = 448$ (with $n = 447$). Then for $b = 462$, the smallest Fermat pseudoprime is $n = 949$. There are a handful more examples of bases $b \leq 1024$ where the first Fermat pseudoprime for base $b$ is $\geq 100$, but they are ``rare."
Let $S$ be the ordered set, indexed by $b \geq 2$, whose elements are the smallest Fermat pseudoprime for base $b$. My instinct is that $S$ is unbounded. In other words, given any $M\in \mathbb{R}_{> 0}$ there exists a pair $(n,b)$ such that $n \geq M$ and $n$ is the smallest Fermat pseudoprime for the base $b$.
I suspect the proof involves Theorem 1 of this paper, and constructing an infinite family of numbers $\{n_I\}$ such that there is only one base $b\:\left(\mathrm{mod}\: n_I\right)$ (corresponding to the "trivial" base $b=1$) where $n_I$ is a Fermat pseudoprime for $b$. It's likely that I am thinking too hard about this, so I am wondering if someone can provide insight about the unboundedness of $S$.
$S$ is unbounded : For a positive integer $n\ge 2$ , just consider the base
$$b:=lcm(2,\cdots ,n)$$
Then , the smallest pseudoprime must exceed $n$ because for $1<m\le n$ there is a prime $p$ both dividing $m$ and $b$ , so $b^{m-1}\equiv 1\mod m$ is impossible.