Consider, for instance, the categories of $C^k$-manifolds, where $k=0,1,2,...,\infty,\omega$. ($C^\omega$ means real analytic.) Are these categories pairwise non-equivalent?

Of course, the obviuos forgetful functor is not an equivalence, because it is not full: for $k<l$, there are $C^k$ functions which are not $C^l$.

I was just thinking about possible ways of proving non-equivalence of categories and came up with this question. For example, one can prove non-equivalence by showing that limits or colimits of certain shape exist in one category but not in the other. I don't know whether one can deal with manifolds in this way.

What I can prove is that the category of *complex*-analytic manifolds is not equivalent to any of the above.

This follows from the fact that any real manifold other than the point (which is the terminal object) has non-trivial automorphisms. For $C^k$-manifolds with $k=0,1,...,\infty$ this is very easy to show, as there is a $C^\infty$-diffeomorphism of the ball $|x|<1$ which is the identity near its boundary sphere $|x|=1$. (One can construct such a diffeomorphism uning bump functions.) This does not work for the real analytic case. However, by a theorem of Grauert, for any real analytic manifold $M$ there is a closed analytic embedding $i$ of $M$ into some $\mathbb R^n$. Then one can choose a non-zero vector tangent to $i(M)$ at some point, extend it to a constant vector field along $i(M)\subset\mathbb R^n$ and project this vector field orthogonally to get a bounded vector field on $M$ which is not identically zero. As $M$ is complete in the induced metric (because $\mathbb R^n$ is complete and $i$ is closed), the flow of this vector field will be globally defined and thus it will contain a non-trivial analytic self-diffeomorphism of $M$.

On the other hand, there are lots of complex analytic manifolds with no non-trivial automorphisms. For example, a generic compact Riemann surface of genus $\geq 3$ is known to have this property.