First off: for my purposes, let $\sf B$ be the category of finite sets with bijections, and ${\sf B}_n$ the subcategory of sets with cardinality $n$, and define a combinatorial species to be a functor $F:{\sf B}_n\to{\sf B}$. This captures the idea of building "structures" on $n$-sets. Write $[n]=\{1,\cdots,n\}$. Every permutation $[n]\to[n]$ induces a corresponding map $F[n]\to F[n]$, so we always have that $F[n]$ is an $S_n$-set.
Can we have nonisomorphic species $F\not\cong G$ for which $F[n]\cong G[n]$ as $S_n$-sets?
In practice, the only way I've been able to prove two species are nonisomorphic (e.g. $S_k\times\binom{-}{k}$ versus ${\rm injhom}(k,-)$) is to show that the corresponding sets $F[n]$ and $G[n]$ are nonisomorphic as $S_n$-sets - eventually I figured out this was the key fact I was using to prove nonisomorphism - but I haven't thought of anything else to accomplish this.
If the isomorphism type of $F[n]$ as an $S_n$-set was enough to determine $F$ it would mean we could "build" a unique (up to isomorphism) species for every $S_n$-set, and this strikes me as impossible because of how "unbased" a generic set $X$ is, although this is a fuzzy intuition at the moment.
Note this is beyond finding inequivariant $S_n$-sets with the same cycle index polynomial, since the isomorphism $F[n]\cong G[n]$ as $S_n$-sets will imply identical cycle index polynomials.
Ideas or comments?
Since the category of finite sets with bijections is equivalent to the permutation groupoid $\coprod_{n \geq 0} S_n$, two combinatorial species are isomorphic if and only if the corresponding $S_n$-sets are isomorphic for all $n$. If you do not believe this, here is a direct argument: For each $n$, let $\alpha_n : F([n]) \to G([n])$ be an isomorphism of $S_n$-sets. We claim that for every bijection $f : [n] \to [m]$ we have $\alpha_m \circ F(f) = G(f) \circ \alpha_n$ ($*$). Since $f$ is bijective, we have $n=m$. But since $\alpha_n$ is an isomorphism of $S_n$-sets, we have $\alpha_n \circ F(f) = G(f) \circ \alpha_n$, and we are done. Now if $X$ is any finite set, choose some bijection $f : X \to [n]$ and define $\alpha_X : F(X) \to G(X)$ by $\alpha_X := G(f)^{-1} \circ \alpha_n \circ F(f)$. Using ($*$) one checks that $\alpha_X$ is (1) independent from $f$, (2) natural in $X$.