Let $\{(U_\alpha,\phi_\alpha)\}_{\alpha\in\mathcal{A}}$ and $\{(U_\beta,\phi_\beta)\}_{\beta\in\mathcal{B}}$ be two smooth (that is $\mathcal{C}^{\infty}$) atlases on a topological manifold $M$. By definition the $\mathcal{C}^{\infty}$-structures defined by these two atlases are equivalent if their union $\{(U_\alpha,\phi_\alpha)\}_{\alpha\in\mathcal{A}\cup\mathcal{B}}$ is also a smooth atlas, that is, the transition maps between the charts of different atlases are smooth. It can be shown, that this definition, is equivalent to saying that two $\mathcal{C}^{\infty}$-structures are equivalent if they determine the same set of smooth functions $f:M\rightarrow\mathbb{R}$.
Here is an example of two non-equivalent $\mathcal{C}^{\infty}$-structures:
Consider the topological manifold $\mathcal{N} = \mathbb{R}$ equipped with the $\mathcal{C}^{\infty}$-structure determined by the $\mathcal{C}^{\infty}$-atlas consisting of the single chart $(\mathbb{R}, Id)$.
Next consider $\mathbb{R}$ equipped with the $\mathcal{C}^{\infty}$-structure determined by the $\mathcal{C}^{\infty}$-atlas consisting of the single chart $(\mathbb{R}, \phi)$ where $\phi(x) = x^{3}$.
We can now show, that each one of the above $\mathcal{C}^{\infty}$-atlases (and thus: their $\mathcal{C}^{\infty}$-structures) are non-equivalent (because, for example, the change of coordinates: $[Id\circ\phi^{-1}](x)=x^{1/3}$ is not smoothly differentiable at $0$), therefore, the $\mathcal{C}^{\infty}$-structures determined by them are different.
From your comment, it appears that your real question is:
Suppose that $X$ is a smooth manifold of positive dimension. Is there a self-homeomorphism $f: X\to X$ which is not a diffeomorphism?
(Given such $f$, the pull-back of the smooth atlas on $X$ via $f$ defines a smooth structure on the topological manifold underlying $X$ which is not equivalent to the original smooth atlas.)
Here is a general construction of $f$. Let $X$ be a smooth manifold of dimension $n\ge 1$, let $\phi_\alpha: U_\alpha\to R^n$ be one of the charts (which I assume to be surjective), where $U_\alpha\subset X$ is open. Now, consider the closed unit ball $B=B(0,1)\subset R^n$ with spherical coordinates $(r,\theta), r\in [0,1], \theta\in S^{n-1}$. Define the self-homeomorphism
$$
h: B\to B, h(r,\theta)=(\sqrt{r},\theta).
$$
I leave it to you to verify that $h$ is not differentiable at the origin (it does not even have the directional derivative at the origin along any nonzero vector). Transplant $h$ to $X$ via the formula
$$
h_\alpha = \phi^{-1}_\alpha \circ h \circ \phi_\alpha.
$$
Set $B_\alpha:= \phi_\alpha^{-1}(B)$.
Then $h_\alpha$ is a self-homeomorphism
$$
B_\alpha\to B_\alpha.
$$
The map $h$ restricts to the identity map of the boundary of $B$, hence, $h_\alpha$ restricts to the identity map of the boundary of $B_\alpha$. Thus, extend $h_\alpha$ by the identity to $X\setminus B_\alpha$. I leave it to you to verify that the resulting map $f: X\to X$ is a homeomorphism and that it is not a diffeomorphism (since it is not differentiable at $\phi_\alpha^{-1}(0)$).
Edit 1. I think, I understood your difficulty. When we say that a map between two subsets of $R^n$ is a homeomorphism, it is important to specify both domain and codomain of the map. But when we talk about differentiability of the same map at some point $p$ in the interior of the domain, we by default extend the codomain to be the entire $R^n$. For instance, the definition of the directional derivative
$$
D_vf(p)=\lim_{t\to 0} \frac{f(p+tv) - f(p)}{t}
$$
requires us to work with vector-valued functions, whose codomains are the entire $R^n$.
Edit 2. Ok, since it is still unclear, let's verify that by map $f$ is not differentiable at the point $p=\phi_\alpha^{-1}(0)$. Consider the open subset $V:=\phi_\alpha^{-1}(int B)\subset X$. The map $f$ that I defined sends $V$ to itself. The map $\psi=\phi:= \phi_\alpha|_V: V\to int B$ is a chart in the smooth atlas of $X$. Consider the composition
$$
(\psi\circ f \circ \phi^{-1})|_{int B}= (\phi_\alpha \circ f \circ \phi_\alpha^{-1})|_{int B}.$$
By the very definition of the map $f$, the above composition equals
$$
(\phi_\alpha \circ \phi_\alpha^{-1}\circ h \circ \phi_{\alpha}\circ \phi_\alpha^{-1}) |_{int B}= h|_{int B}.
$$
If $f$ were differentiable at $p$ then this composition would have been differentiable at $0$ as well. However, as I noted above, $h$ is not differentiable at $0$. Thus, $f$ is not differentiable at $p$.
Best Answer
Suppose $\varphi_\beta\circ\varphi_\alpha^{-1}$ isn't smooth. Then one of $x_i\circ\varphi_\beta$ isn't smooth in $(\varphi_\alpha)_{\alpha\in A}$, while it's clearly smooth in $(\varphi_\beta)_{\beta\in B}$. Choose a sufficiently small compact neighborhood $K$ of a point in which it isn't smooth. Then $x_i\circ\varphi_\beta|_K$ can be smoothly extended to the whole $M$ and is the function you want.