Why is HAMILTONIAN CYCLE so different from PERMANENT?
The following is a proof over any ring of characteristic zero that the Hamiltonian cycle polynomial is not a polynomial-size monotone projection of the permanent. The basic idea is that monotone projections of polynomials with nonnegative coefficients lead to the Newton polytope of one being an extended formulation of the Newton polytope of the other, and then applying the recent lower bounds on extended formulations.

Let $f(x_1,\dotsc,x_n)$ and $g(y_1,\dotsc,y_m)$ be polynomials with nonnegative coefficients (as is the case here). Suppose that $f$ is a monotone projection of $g$ under the assignment $\pi$ (following the notation of the question). Under $\pi$, each monomial of $g$ gets mapped either to 0 (iff one of its variables gets mapped to 0) or to a monomial of $f$: there can't be any cancellations because of non-negativity. 

Let $New(f)$ denote the Newton polytope of $f$, and similarly for $New(g)$.

Claim: there is an extended formulation for $New(f)$ in $\mathbb{R}^m$, using $\leq n+m$ variables, and a number of constraints that is at most $n+m$ plus the number of constraints defining $New(g)$.

Here's how: Let $e_1,\dotsc,e_m$ be the coordinates on $\mathbb{R}^m$ (in which $New(g)$ lives; viz., an integer point in $\mathbb{R}^m$ with coordinates $(e_1,\dotsc,e_m)$ corresponds to the monomial $y_1^{e_1} \dotsb y_m^{e_m}$). For those $i$ such that $\pi(y_i)=0$, intersect $New(g)$ with $\{e_i = 0\}$ (since only monomials that do not involve $y_i$ can contribute to the projection); this adds at most $m$ additional constraints. Let $P$ denote the resulting polytope. Then $\pi$ induces a linear map $L_{\pi}\colon \mathbb{R}^{m} \to \mathbb{R}^{n}$, such that $L_{\pi}(P) = New(f)$. This last part follows from the lack of cancellations. Thus we get an extended formulation for $New(f)$ by taking $n+m$ variables, the constraints for $P$ on the $m$ variables, and the constraints defining $L_{\pi}$ (of which there are at most $n$, one for each $x_i$). QED Claim

Now take $f$ to be the $n$-th Hamilton cycle polynomial and $g$ to be the $m$-th permanent, and suppose that $f$ is a monotone projection of $g$. The Newton polytope of the permanent (and determinant, incidentally), is the cycle cover polytope. This polytope is easily described by the "edge" variables $e_{ij}$ and the $m$ equations stating that every vertex has degree exactly 2.

The Newton polytope of the Ham. Cycle polynomial is the Hamiltonian cycle polytope (surprise, surprise). But this polytope is the TSP polytope, which requires $2^{n^{\Omega(1)}}$ equations to describe any extended formulation of, which, when $m$ is subexponential, contradicts the small extended formulation given by the cycle cover polytope and $L_{\pi}$ as above.

(Note that this argument fails if $f$, $g$, or $\pi$ can have negative coefficients, as then there can be cancellations, so $L_{\pi}$ need not map onto $New(f)$.)

It's interesting to note that the geometry of these polytopes is closely related to the fact that matching is in $\mathsf{P}$ while Hamilton cycle is $\mathsf{NP}$-complete, but I think the above reasoning shows how the geometric structure here really can add something beyond that complexity classification.