As [this article](https://en.wikipedia.org/wiki/Eulerian_path),I think we want to find all of the *Eulerian path*.But *Mathematica* have no such function to do this directly.So I will delete the edge `1 <-> 2` first,then use `FindEulerianCycle` like follow:
###Make a intermediate graph without edge `1 <-> 2`:

    pts = {{0, 0}, {1, 0}, {1, 1}, {0, 1}, {1/2, 1 + Sqrt[3]/2}};
    g = EdgeDelete[
      Graph[{1 <-> 2, 2 <-> 3, 3 <-> 4, 1 <-> 3, 1 <-> 4, 2 <-> 4, 
        4 <-> 5, 3 <-> 5}, VertexCoordinates -> pts, 
       VertexLabels -> "Name"], 1 <-> 2]

[![enter image description here][1]][1]
###Find all of the *Eulerian path*

    Prepend[#, 1 <-> 2] & /@ FindEulerianCycle[g, All]

[![enter image description here][2]][2]

PS:I found the vertex $3$ and $4$ is completely equivalent.So you can find another $6$ path.(Actually I think this is a bug of `FindEulerianCycle` which can find another $6$ path.)
  [1]: https://i.sstatic.net/rJScR.png
  [2]: https://i.sstatic.net/oBG9a.png