Question: Consider the proper class of all $n$-dimensional smooth manifolds. If we take the equivalence classes where two manifolds are identified if there exists a diffeomorphism between them, is this collection of equivalence classes a set?
Remark: I do not assume my manifolds to be Hausdorff nor second countable. If the answer depends on those assertions (Edit: it most definitely does) I would like to hear about the difference.
Remark 2: As Omar pointed out, there may be a problem with the various long lines. To make the question slightly more tractable, I would be (mostly) satisfied if there is a statement even ignoring the smooth structure and consider the case of topological manifolds and homeomorphisms.
Motivation: for something that I am working on I need to consider the collection of all $n$-dimensional smooth manifolds satisfying "property $X$". Unfortunately property $X$ is diffeomorphism invariant, so most definitely this collection is not a set, which invalidates many constructions (I want to build a manifold out of this collection; if the collection is a proper class then even on the set level the thing that I constructed will be a proper class, instead of a set) or at least forces me to rethink how this constructions ought to go. Fortunately for my argument it suffices that I have one object in each diffeomorphism class in my collection.