Grow a square in the hyperbolic plane until its vertex angles become $\pi/5$. Assuming that the constant Gaussian curvature of our hyperbolic plane is $-1$, the sides of the resulting hyperbolic square will have length $\operatorname{arccosh}(\cot(\pi/10)^2)\approx 2.94$, while its main diagonals will have length $2\operatorname{arccosh}(\cot(\pi/10))\approx 3.58$.
Tile the hyperbolic plane with copies of this square, so that ten tiles meet at each vertex. Then identify each square with its four king-wise neighbors by reflection in their common edge. The resulting quotient will be a single square, but with its boundary converted into a mirror edge and with its four corners converted into kaleidoscopic points of order 5, lying along that edge. Viewed as an orbifold, the result is thus ${\ast}5555$, using Conway's signature scheme for orbifolds. (There is actually a one-parameter family of orbifolds with this signature, since we can replace the initial square with a rectangle of any aspect ratio. But let's focus on the most symmetric case, which starts out with a square.)
We now do one more identification: we identify the remaining square with the result of rotating it by $\pi$ around its center. This converts the center of the square into a cone point of order 2 and identifies opposite points along the mirror edge; so we now have the orbifold $2{\ast}55$. Can this orbifold be smoothly and isometrically embedded in Euclidean 3-space? (Once again, we focus on the most symmetric orbifold of this type, in which the two kaleidoscopic points are antipodal along the mirror edge.)
It is well known, of course, that the entire hyperbolic plane cannot be isometrically embedded in 3-space with even $C^2$ smoothness, though a $C^1$ embedding is possible. But we are dealing here with only a small piece of that plane.
Dini's surface provides a real-analytic, isometric embedding of a good-sized region of the hyperbolic plane. That region includes hyperbolic disks of any finite radius, though big disks will be wrapped many times, very tightly around the screw axis of Dini's surface. The orbifold ${\ast}5555$ fits inside a disk of radius $\operatorname{arccosh}(\cot(\pi/10))\approx 1.79$, so it can be embedded isometrically and real-analytically; but that radius is already large enough to require quite a bit of wrapping around the axis.
The orbifold $2{\ast}55$ has a cone point, which is a new challenge. Take two rays in the plane that make an angle of $\pi/6$, and revolve one of them around the other. The resulting surface of revolution is a cone whose tip is a cone point of order 2. This cone has Gaussian curvature zero everywhere (excluding its tip); but there is a unique way to make it flare out, so as to achieve constant Gaussian curvature $-1$, while remaining a surface of revolution. The result is a classical surface of constant negative curvature, referred to as being "of conical type". Indeed, the better-known pseudosphere is the limit of such surfaces as the angular width of the cone goes to zero while the cone's tip simultaneously goes off to infinity.
It is tempting to try to embed $2{\ast}55$ isometrically by using the flared cone whose cone point has order 2. Unfortunately, the hyperbolic distance from the tip of that flared cone, along a meridian, to the boundary edge beyond which it cannot be continued is only $1.317$ (calculated using NDSolve in Mathematica). Since the distance, in the orbifold $2{\ast}55$, from the cone point to the mirror edge varies between $0.81$ and $1.79$, the two "ears" of the orbifold $2{\ast}55$ will fall off the edge of the flared cone. (The analogous orbifold $2{\ast}33$ has a max radius of only $\operatorname{arccosh}(\cot(\pi/6))\approx 1.15$, so it would actually fit onto the flared cone. But the radius of $2{\ast}44$ varies between $0.76$ and $\operatorname{arccosh}(\cot(\pi/8))\approx 1.53$, so its two "ears" would also fall off.)
Given that the flared cone is too small for $2{\ast}55$, is there some other way to embed that orbifold in 3-space isometrically and real-analytically, presumably using some surface that is not a surface of revolution?