Lemma: In a polyhedron of this type with polygons $A$ and $B$ sharing an edge $e$, the two other polygons meeting $e$ must have the same number of sides.

Proof: By local symmetry reflecting through the perpendicular bisector of $e$, the angles are equal.

Sergei Ivanov proved the same lemma in the comments.

Since 5 is odd, all of the polygons around a pentagon must have the same number of sides, since you can't have a nonconstant alternating sequence. So, the only possibilities are that all polygons are pentagons, or that each pentagon is surrounded by hexagons, and each of these hexagons is surrounded by 3 pentagons and 3 hexagons. In the latter case, attaching pentagonal pyramids to each pentagon extends each hexagon into an equilateral triangle, producing a polyhedron whose faces are equilateral triangles with 5 meeting at a vertex, an icosahedron, so the original was a truncated icosahedron.

Note that if you have equilateral triangles and squares meeting 4 to a vertex, then there are two possibilities for 3 squares and 1 triangle at a vertex, one with cubic symmetry and one with only dihedral symmetry with a belt which is an octagonal prism.

By contrast, if you require that there are 3 triangles meeting at a vertex, but drop the regularity assumption, you get a family of disphenoids, which generically have the Klein 4-group as symmetries and no reflective symmetry.

Lemma: In a polyhedron of this type with polygons $A$ and $B$ sharing an edge $e$, the two other polygons meeting $e$ must have the same number of sides.

Proof: By local symmetry reflecting through the perpendicular bisector of $e$, the angles are equal.

Sergei Ivanov proved the same lemma in the comments.

Since 5 is odd, all of the polygons around a pentagon must have the same number of sides, since you can't have a nonconstant alternating sequence. So, the only possibilities are that all polygons are pentagons, or that each pentagon is surrounded by hexagons, and each of these hexagons is surrounded by 3 pentagons and 3 hexagons. In the latter case, attaching pentagonal pyramids to each pentagon extends each hexagon into an equilateral triangle, producing a polyhedron whose faces are equilateral triangles with 5 meeting at a vertex, an icosahedron, so the original was a truncated icosahedron.

Note that if you have equilateral triangles and squares meeting 4 to a vertex, then there are two possibilities for 3 squares and 1 triangle at a vertex, one with cubic symmetry and one with only dihedral symmetry with a belt which is an octagonal prism.

By contrast, if you require that there are 3 congruent triangles meeting at a vertex, but drop the regularity assumption, you get a family of disphenoids, which generically have the Klein 4-group as symmetries and no reflective symmetry. These are related to ideal hyperbolic tetrahedra.

Lemma: In a polyhedron of this type with polygons $A$ and $B$ sharing an edge $e$, the two other polygons meeting $e$ must have the same number of sides.

Proof: By local symmetry reflecting through the perpendicular bisector of $e$, the angles are equal.

Sergei Ivanov proved the same lemma in the comments.

Since 5 is odd, all of the polygons around a pentagon must have the same number of sides, since you can't have a nonconstant alternating sequence. So, the only possibilities are that all polygons are pentagons, or that each pentagon is surrounded by hexagons, and each of these hexagons is surrounded by 3 pentagons and 3 hexagons. In the latter case, attaching pentagonal pyramids to each pentagon extends each hexagon into an equilateral triangle, producing a polyhedron whose faces are equilateral triangles with 5 meeting at a vertex, an icosahedron, so the original was a truncated icosahedron.

Note that if you have equilateral triangles and squares meeting 4 to a vertex, then there are two possibilities for 3 squares and 1 triangle at a vertex, one with cubic symmetry and one with only dihedral symmetry with a belt which is an octagonal prism.

By contrast, if you require that there are 3 triangles meeting at a vertex, but drop the regularity assumption, you get a family of disphenoids, which generically have the Klein 4-group as symmetries and no reflective symmetry.

Lemma: In a polyhedron of this type with polygons $A$ and $B$ sharing an edge $e$, the two other polygons meeting $e$ must have the same number of sides.

Proof: By local symmetry reflecting through the perpendicular bisector of $e$, the angles are equal.

Sergei Ivanov proved the same lemma in the comments.

Since 5 is odd, all of the polygons around a pentagon must have the same number of sides, since you can't have a nonconstant alternating sequence. So, the only possibilities are that all polygons are pentagons, or that each pentagon is surrounded by hexagons, and each of these hexagons is surrounded by 3 pentagons and 3 hexagons. In the latter case, attaching pentagonal pyramids to each pentagon extends each hexagon into an equilateral triangle, producing a polyhedron whose faces are equilateral triangles with 5 meeting at a vertex, an icosahedron, so the original was a truncated icosahedron.

Note that if you have equilateral triangles and squares meeting 4 to a vertex, then there are two possibilities for 3 squares and 1 triangle at a vertex, one with cubic symmetry and one with only dihedral symmetry with a belt which is an octagonal prism.

By contrast, if you require that there are 3 triangles meeting at a vertex, but drop the regularity assumption, you get a family of disphenoids, which generically have the Klein 4-group as symmetries and no reflective symmetry.