There's an obvious operation on the category of graded rings, given by "scaling up," multiplying the grading of every element by some fixed constant.
Does anyone know of an operation on the level of spaces that will do this to the cohomology rings?
There's an obvious operation on the category of graded rings, given by "scaling up," multiplying the grading of every element by some fixed constant. Does anyone know of an operation on the level of spaces that will do this to the cohomology rings? 


Even if you don't ask for the ring structure to be preserved, this seems quite unlikely, at least if you require the functor to do the same "scaling up" operation to maps on cohomology induced by maps of spaces. I believe all the Goodwillie derivatives of such a functor would be zero, whereas "geometric" operations typically have some nontrivial radius of analyticity. Edit: For the argument I was thinking of, I need to assume that π_{1}(F(any sufficiently highly connected space)) = 0, so that for instance F(•) = •. 


There isn't one. The scaling operation that you describe (which would have to be an odd scale factor, by the way) doesn't commute with suspension. Cohomology is a "suspensionpreserving" functor (there's probably some fancy way of saying this using triangulated categories) so when you consider a cohomology theory you should really think of the target as "graded algebras over the coefficient ring with a suspension operation" only that's not quite as snappy as "graded rings" so we tend to be a bit sloppy about saying it. Edit: This answer was a bit of a "throwaway" answer since the question felt like idle speculation (I'm a bit embarrassed that it got so many votes!). With a little more thought I'd've concentrated on the fact that the target category isn't graded rings, or even graded algebras, or even graded algebras with suspension, but there's the action of the cohomology operations to take into account. Simply regrading everything will completely mess up those actions. Suspensions will come into play, though, when one considers maps between spaces and I suspect that one could show that all such maps were nullhomologous. I like Tyler's answer best. I'm voting for that one, and I recommend it be accepted as the answer. 


To get from spaces to graded rings you go via dgrings. That is, Spaces → dgRings → graded Rings. where the first arrow is the functor taking simplicial cochains, and the second is the functor taking cohomology. So any "scaling up" of spaces should exist in dgRings. But how do you scale up a cochain complex? You can't because because that would break the degree of the codifferential. So I don't think that there is a sensible "scaling up" for spaces. Nice idea though, I don't know how you could classify functors that act on graded rings that can be lifted to spaces. An obvious example might be that of setting all groups of degree greater than i say to 0. That would correspond to filling in things in the space to kill the cohomology groups (I don't know the correct terminology off the top of my head). 


There is no such operation. Here is one reason why. Take your space to be CP^{2}. Its cohomology ring is a truncated polynomial algebra ℤ[x]/(x^{3}), where x is an element in degree 2. Suppose we had such a shift functor S on spaces that multiplies cohomology degrees out by n. Then S(CP^{2}) would have mod2 cohomology ring a truncated polynomial algebra ℤ/2[y]/(y^3) on a generator y in degree 2n. This violates the Hopf invariant 1 conjecture, proved by Adams. There are no spaces whose cohomology is such a truncated polynomial ring ℤ/2[y]/(y^3) on a generator in degree k unless k is 1 (real projective 2space), 2 (complex projective 2space), 4 (quaternionic projective 2space), or 8 (something associated to the octonions that I feel uncomfortable describing as projective space). If n is not 2 or 4, you immediately get a contradiction here. If n is 2 or 4, you could apply your shift operator multiple times and get a contradiction. 


How about we change the question to this: for what kind of spaces X are there a family of continuous maps ψ^{k}: X>X which act on cohomology by "scaling up"? Here is a famous and bewildering example of something sort of like this: As you can read in the notes of Sullivan's 1970 MIT course (see esp. chapter 5), if you have an algebraic variety X defined over Q(the rationals), then you get an action of G(=absolute Galois group of Q) on a space X_{et}=the etale homotopy type of X. If X is a Grassmanian variety, then * X_{et} has the homotopy type of the usual complex Grassmannian (up to a "profinite completion"), * G acts on the profinite cohomology of X_{et} through its abelianization G_{ab}=Z^{*}(=the units of the profinite integers), * this action of G_{ab} on cohomology is by scaling. I don't know what the motivation of the original question is, so I don't know if this kind of thing has any relation for what you want. (I really posted this to advertise the Sullivan notes, which everybody should read.) 

