I'm reading a book, and this seems to be true, but they don't say why, and I'm not sure.
I'm a little confused. While it is true that the cotangent space of a manifold naturally carries the structure of a smooth manifold (it's just a finite dimensional vector space) this is a strange question. I believe what you meant to ask is whether or not a diffeomorphism induces a linear isomorphism on the cotangent spaces. This is in fact true. In particular, let's suppose that we have a diffeomorphism $\displaystyle f:M\to N$. I think you know then that the derivative $\displaystyle df$ induces a linear isomorphism $\displaystyle T_xM\to T_{f(x)}N$ for each point $\displaystyle x\in M$. Dualizing this, you get an isomorphism $\displaystyle df^\ast(x):T_x^\ast N\to T_x^\ast M$, which is the desired isomorphism.
small addendum: linear maps are always differentiable, and if a linear map is invertible, it's inverse is also a linear map, and thus differentiable, so a linear isomorphism is a diffeomorphism (not a very interesting one, though).
I'd like to add, that if your manifold has a Riemannian structure there is a covariant way to do this. Namely, take the isomorphism $\displaystyle df(x):T_xM\to T_{f(x)}N$ and consider the identification $\displaystyle T_xM\cong T_X^\ast M$ given by your metric.