In how many ways can an iterated tangent bundle (T^k)M be viewed as a fibre bundle over (T^(k-1))M? Let M be a smooth manifold. The double tangent bundle, TTM,can be viewed as a fibre bundle over TM in two ways, with the projection maps given by T_πM (i.e. the derivative of the projection from TM to M) and π_TM (i.e. the standard projection onto TM). There is also a canonical involution K:TTM->TTM, which basically flips the inner two coordinates. It turns out to be a diffeomorphism and a natural transformation from T^2 to itself. In fact, Tπ_M and π_TM are related through composition with K.
My question is, what happens if you keep taking higher and higher tangent bundles? Evidently, you should have more ways to write them as fibre bundles over the lesser tangent bundle. Intuitively, to me at least, it feels that there are k ways to view (T^k)M as a fibre bundle over (T^(k-1))M: inductively, there is the derivative of all the previous projection maps, and the canonical way. Are there any others?
Is there always a diffeomorphism (or whatever the suitable notion is here) that will take one projection map to the other as in the case of the canonical involution in k=2? If not, what goes wrong, and does it have any significance?
 A: AMENDED POST:  Marco Gualtieri points out a fundamental mistake I made. 
I retract my assertion that there infinitely many such maps.
It looks to me there is a  surjective map $T^n M \to T^{k}M$ for each 
monotone injective map $\lbrace 1,\dots,k\rbrace  \to \lbrace 1,\dots, n\rbrace$.

ORIGINAL POST: I believe that "infinitely many" is the answer.
We can think of $TM$ as the effect of applying an equivalence relation (two paths 
are equivalent if they have equal one jets at the origin) on the space of smooth paths $\Bbb R\to M$, which I'll denote as $M^{\Bbb R}$.  Similarly, the $n$-fold iterated tangent bundle $T^nM$ is given by a similar equivalence relation on $M^{\Bbb R^n}$.  For any linear embedding $\Bbb R^k \to \Bbb R^n$ there will be a restriction map $T^nM \to T^kM$.
This gives infinitely many maps. 
Similarly, for  any linear surjection $\Bbb R^n \to \Bbb R^k$ one gets an embedding $T^kM \to T^nM$. 
If we use only coordinate inclusions and coordinate projections, then the structure you get is a simplicial object $X.$ with $X_k = T^{k+1}M$ augmented over $M$.  This is because $M \mapsto TM$ is a triple (comonad).  I've always felt that one could make sense of the de Rham complex from this point of view (somehow). 
A: If we use the notation $(TM, p_M, M)$ for the tangent bundle of any manifold $M$, then you are right to think that $T^{\ k}M$ has $k$ natural vector bundle structures over $T^{\ k-1}M$ and so on down to $M$, making a diagram which is a $k$-dimensional cube.  Such a structure is a $k$-fold vector bundle (See articles by Kirill Mackenzie) and $T^{\ k} M$ is a particularly symmetrical one.  An easy way of writing the $k$ bundle maps would be 
$$T^{\ a}(p_{T^{\ b} M}) $$
for $a+b+1 = k$, where this means that we are taking the $a$-th derivative of the tangent bundle projection $T\ (T^{\ b}M)\to T^{\ b}M$, yielding a map now from $T^{\ k}$ to $T^{\ k-1}$.  
To see the symmetries of $T^{\ k} M$ it is more convenient to describe the functor in a direct way rather than as a $k$-fold composition -- just as you might think of tangent vectors as infinitesimal curves, you can think of points in $T^{\ k} M$ as infinitesimal maps of a unit $k$-cube into $M$. The restrictions to the $k$ faces of the cube (going through the origin) gives your $k$ maps. 
From that point of view, it is clear that you can permute the $k$ coordinate axes and get another map of a cube into $M$, so that the functor $T^{\ k} M$ has a $S_k$ group of natural-automorphisms.
Incidentally, to make the above into a definition of $T^{\ k} $, you could do the following:
Consider the fat point $fp$, which you should think of as a space whose smooth functions form the ring $\mathbb{R}[x]/(x^2)$.  Then the tangent bundle $TM$ can be thought of as the space of maps from the fat point to $M$, i.e. $TM=C^\infty(fp,M)$.  Such maps, by the way, are just algebra homomorphisms from the algebra $C^\infty(M,R)$ to $\mathbb{R}[x]/(x^2)$.  You can check that such a map has two components $f_0 + f_1 x$, and that $f_0$ is a homomorphism to $\mathbb{R}$ defining a maximal ideal (i.e. a point $p$ in $M$) and $f_1$ defines a derivation (i.e. a vector at $p$).   
In precisely the same way you can consider a cubical fat $k$-point $kfp$ with functions 
$$\mathbb{R}[x_1,...,x_k]/(x_1^2,...,x_k^2)$$
and then define $T^{\ k} M = C^\infty(kfp,M)$. Then you can see the symmetries as automorphisms of the above algebra, and the $k$ maps you ask about as homomorphisms $C^\infty(kfp)\to C^\infty\big((k-1)fp\big).$
There are probably more subtle things to be said about these higher iterated bundles but I hope the above is at least correct.
A: More background to Marco Gualtieri's answer can be found in chapter VIII of 
here, which classifies all product preserving functors on the category of smooth manifolds and smooth mappings, and all natural transformations between them. This result is partially extended to the category of infinite dimensional smooth manifolds in section 31 of here
