Show the following graph is planar I am trying to show that a graph is planar. Possibly the simplest method I have found is to show the graph can be drawn on a page (i.e. in the plane) without any edges crossing. So, can I assume that if I am given a graph that has edges crossing I can simple move the vertices around to obtain a version such that the edges are not crossing (if such an arrangement exists)?
It may be better to show an example of what I am thinking. Perhaps someone can confirm that what I have done is permutable. Refer to the figure below were I start with the left graph and end with the right graph.

Alternatively, it seems as though one can show a graph is planar if it can be embedded in a disk. I believe the following figure shows such an embedding in a disk

 A: Your numbering scheme is confusing. Determining whether two graphs are isomorphic is difficult for humans to do; one of the reasons we label vertices is to help the reader identify that two graphs are isomorphic at a glance. Your numberings are misleading in that regard, because the "isomorphism" sending vertex-$1$-on-the-left to vertex-$1$-on-the-right, through to vertex-$6$-on-the-left to vertex-$6$-on-the-right, is not a graph homomorphism.
The following diagram demonstrates the planar structure and it is easy to see that the two graphs are isomorphic.

I think in your disk embedding you have missed the $4 \to 6$ edge.

Alternative way: Kuratowski's theorem states that a graph is planar iff it contains no subdivision of $K_{3,3}$ and no subdivision of $K_5$.
Clearly the graph contains no subdivision of $K_{3,3}$, because it already has six vertices, and the only six-vertex subdivision of $K_{3,3}$ is $K_{3,3}$ itself. But your graph isn't $K_{3,3}$ because it contains a vertex of degree $4$.
Also it contains no subdivision of $K_5$ because that could only be obtained by adding a single vertex to an edge of $K_5$; the resulting graph would have at least three vertices of order $4$, but your graph only has one.
