Are there different combinatorial species with the same symmetry type? 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?
 A: 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 $X$.
A: Here are my ideas and comments, explained for the less initiated reader. 
Note that the expression "transport of structures" is misused in the theory of species. The thing transported by functoriality is only and only the subjacent symmetry of a structure - and this is why we are looking to permutation groups and cycle indices. 
The first two permutation groups with the same cycle index are 8T18 and 8T22 in Maple, so the cycle index does not define a species but the question is beyond this: then what defines a species up to isomorphism ?  As shown in the answer, only the symmetry matters. This is somehow unbelievable given the word "structure" is used by tradition instead of "symmetry".
For example, take the Chvátal graph and an octogon with four marked edges. They have the same subjacent species, and nothing else in common - where it comes to labeling.

