Are all degenerate d-orbitals identical?
Yes, they are identical.
One thing that we don't really teach well with orbitals is thinking about the symmetry of the orbital with respect to the name of the orbital.
$p_{x}$ has the same symmetry as the function $f(x,y,z)=x$.
Likewise, $d_{x^{2}-y^{2}}$ has the same symmetry as $f(x,y,z)=x^{2}-y^{2}$.
What about $d_{z^{2}}$? You should note that $d_{z^{2}}$ is really $d_{2z^{2}-x^{2}-y^{2}}$.
Imagine taking $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$, summing them up, and renormalizing. So the symmetry is like: $z^{2}-x^{2} + z^{2} - y^{2}=2z^{2}-x^{2}-y^{2}$.
What you get is a big lobe of the same phase along the $z$-axis. And around it in the $x$-$y$-plane (but smaller), you see a smear of the opposite phase that's shaped kind of like a torus. That's why it looks different. It's really two of the other looking orbitals put together.
Why don't we just use $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$ instead? Because they're not linearly independent with the 4 other &quot;normal&quot;-looking orbitals. $d_{x^{2}-y^{2}}$ summed with $d_{z^{2}-x^{2}}$ is $d_{z^{2}-y^{2}}$. The 5 conventional orbitals are just the nice linear combinations of the 5 spherical harmonic solutions to the angular part of the Schrodinger equation. See here.
You don't seem to have a problem with $d_{xy}$, $d_{zx}$, $d_{yz}$, and $d_{x^{2}-y^{2}}$ all being degenerate. So hopefully, you can see that $d_{z^{2}}$ is not really different and therefore also of the same energy as the other 4.
EDIT:
Based on comments, it's also important to point out that the pictures of orbitals with smooth surfaces are not real pictures of orbitals. These are boundary surface diagrams where we've drawn the surface that represents a constant probability of finding the electron for that orbital. This basically means that electron densities for $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$ will smear together in the $x,y$-plane (the orbitals are the same phase) to create something that is symmetrical and shaped like a torus.
This analysis extends fully to $f$ orbitals. Just identify the full functional name for each orbital and apply the same analysis here.