This is a complete rewrite of my original answer, combined with my comments on the original question and various other answers.
Suppose we have a finite set $C$ of candidates, and a finite set $E$ of electors (with $|C|>1$ and $|E|>2$, to avoid some degenerate cases). I will assume that we are just modelling what happens in some kind of secret ballot after campaigning has finished, so each elector independently fills in some kind of form. Let $F(C)$ be the set of possible ways to fill in the form. If all candidates are treated equally, then $F(C)$ should be functorial for bijections of $C$. The collection of voter choices is a point $f\in\text{Map}(E,F(C))$.
Some obvious candidates for $F(C)$ are
- $F(C)=C\amalg\{0\}$: the traditional system in which each voter selects a single candidate, or does not vote (represented by $0$).
- $F(C)=P(C)$ (the set of subsets of $C$): the "approval voting" system where each voter indicates which candidates they find acceptable
- $F(C)=\text{Ord}(C)$ (the set of total orders on $C$): the system assumed in the Arrow and Gibbard-Satterthwaite theorems, where each voter has and records a total preference order.
- $F(C)=\text{Pre}(C)$ (the set of preorders on $C$).
It is not hard to come up with other possibilities.
Note that if $|C|=n$ and $N=\{0,\dotsc,n-1\}$ then $\text{Ord}(C)$ can be identified with the set of bijections $N\to C$. It follows that natural maps $\text{Ord}\to F$ biject with elements of $F(N)$. In particular, there are plenty of such maps, but we see from the above examples that often none of them will be injective. It seems reasonable to assume that there is a given nonempty set $R_0(N)\subseteq F(N)$, consisting of the possible form responses that are not obviously incompatible with the preferences $0<1<\dotsb<N-1$. Given this, we can define by functoriality a subset $R(C)\subseteq\text{Ord}(C)\times F(C)$, consisting of pairs $(o,u)$ where the form response $u$ is not obviously incompatible with the ordering $o$.
Now put $e=|E|$ and $$ S_e(F(C)) = \{m\colon F(C)\to \mathbb{N} : \sum_{u\in F(C)} m(u) = e\}. $$ Given $f\in\text{Map}(E,F(C))$ we can put $\mu(f)(u)=|f^{-1}\{u\}|$ to get a point $\mu(f)\in S_e(F(C))$, which is a complete $\text{Aut}(E)$-invariant for $f$. Any fair voting system should be $\text{Aut}(E)$-invariant and so should factor through $\mu$.
I will assume for the moment that we now want to elect a single candidate, so $n=1$ in Tom's notation.
Ideally we might hope for a map $\sigma\colon S_e(F(C))\to C$ such that $\sigma(\mu(f))$ is the successful candidate. Fairness between candidates dictates that this should be equivariant for $\text{Aut}(C)$. However, this is clearly impossible, because there will usually be many points in $S_e(F(C))$ that are fixed by $\text{Aut}(C)$, but there are no such points in $C$. Thus, we need some way to think about breaking ties, even if they are likely to be rare.
The approach of Duggan and Schwartz (http://dx.doi.org/10.1007%2FPL00007177; http://en.wikipedia.org/wiki/Duggan%E2%80%93Schwartz_theorem) is to consider voting systems $\sigma\colon S_e(\text{Ord}(C))\to P'C$, where $P'C$ is the set of nonempty subsets of $C$. The idea is that $\sigma(f)$ will usually be a singleton, but if not, one of the candidates in $\sigma(f)$ will be chosen by some kind of lottery. They show that under minimal assumptions, any such system is manipulable, no matter what the details of the lottery might be. I think that this completely resolves the question for the case $F=\text{Ord}$, at least if we accept the traditional position that manipulability is the key thing to avoid.
So what if $F\neq\text{Ord}$? I have tried to think of other ways of handling ties, but it seems to me that the Duggan-Schwartz framework is optimal. Thus, we should think about $\text{Aut}(C)$-equivariant voting systems $\sigma\colon S_e(F(C))\to P'C$. We then need to define what it would mean for such a system to be manipulable. It seems inescapable that such a definition must involve the notion that some voters prefer some outcomes to some other outcomes. Thus, we must assume that each voter has a preference preorder. Some voters may be completely apathetic, so their preorders will rank all candidates equally, but some other voters may have a total preference order. Any satisfactory system must be able to handle the special case where every voter's preorder is total, so if we can prove an impossibility theorem in that context, then we are done. Now choose a point $r\in R_0(N)$, giving a map $\rho\colon\text{Ord}(C)\to F(C)$. It might happen that all voters choose to fill in their ballot papers by applying $\rho$ to their total preference order. If we can prove an impossibility theorem in this special case, then again we are done. We now have a map $\sigma\circ\rho_*\colon S_e(\text{Ord}(C))\to P'(C)$. The Duggan-Schwartz theorem gives a list of four properties that this map cannot satisfy simultaneously. Because we have assumed stronger symmetry conditions than Duggan and Schwartz, the Citizen Sovereignty and non-Dictatorship conditions are automatic. The Residual Resoluteness condition says that if all voters have the same preference order, except that one voter might swap the top two candidates, then the majority first choice should be elected. In our context this is a condition on $R_0$ and $\sigma$, and it seems like a condition that we should certainly assume. The theorem then says that $\sigma\circ\rho_*$ represents a voting system that is manipulable according to the Duggan-Schwartz definition. This holds for all $r\in R_0(N)$, and that seems like a reasonable sufficient condition to say that $\sigma$ itself is manipulable.
If we want to elect $n$ candidates with $n>1$, we should probably consider a framework similar to that of Duggan and Schwartz, except that $\sigma$ should be a map from $\text{Map}(E,\text{Ord}(C))$ to $P_{\geq n}(C)$, and precisely $n$ candidates should be chosen by lottery if $\sigma$ produces a set of size strictly larger than $n$. It looks to me as though nothing much should change, but I have not tried to work out the details.