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*:


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

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

    MapIndexed[
     Export[ToString@First[#2] <> ".gif", #, "DisplayDurations" -> 0.5] &,
      FoldList[HighlightGraph[#1, #2, GraphHighlightStyle -> "Thick"] &, 
        g1, #] & /@ paths]

<img src="https://i.sstatic.net/iS7G6.gif" height="250"/>
<img src="https://i.sstatic.net/nndCW.gif" height="250"/>
<img src="https://i.sstatic.net/II4tY.gif" height="250"/>
<img src="https://i.sstatic.net/3ObQa.gif" height="250"/>
<img src="https://i.sstatic.net/1Opjb.gif" height="250"/>
<img src="https://i.sstatic.net/RfluO.gif" height="250"/>

**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 cannot find another $6$ path.)


  [1]: https://i.sstatic.net/rJScR.png
  [2]: https://i.sstatic.net/oBG9a.png