In a "coordinate transformation" (change-of-coordinates transformation), if you start with a basis $\beta=[\mathbf{v}_1,\ldots,\mathbf{v}_n]$, then because the transformation is one-to-one, the image of $\beta$ must be linearly independent, and hence a basis. Thus, the image of the transformation contains a basis, and so will necessarily be onto. Thus, by virtue of being one-to-one on a finite dimensional space, it must necessarily be onto as well, and thus will be invertible.
Or you can view a "change-of-coordinates" transformation as a way of expressing vectors in one basis, $\beta$, in terms of vectors in another basis, $\gamma$; if you then write out what it would mean to express the vectors in $\gamma$ in terms of $\beta$, and you compose the resulting two transformations, you get the elements of $\beta$ expressed in terms of $\beta$ (and composing the other way, the elements of $\gamma$ in terms of the vectors of $\gamma$). Because each vector can be expressed in a unique way in terms of the vectors of a basis, the only way to express the vectors of $\beta$ in terms of the vectors of $\beta$ is by the identity transformation: $\begin{align*} \mathbf{v}_1 &= 1\mathbf{v}_1 + 0\mathbf{v}_2 + \cdots + 0\mathbf{v}_n\\ \mathbf{v}_2 &= 0\mathbf{v}_1 + 1\mathbf{v}_2 + \cdots + 0\mathbf{v}_n\\ &\vdots\\ \mathbf{v}_n &= 0\mathbf{v}_1 + 0\mathbf{v}_2 + \cdots + 1\mathbf{v}_n. \end{align*}$ so that the composition is the identity. Composing them the other way also gives the identity, so that shows the original change-of-coordinates transformation must be invertible.