Here's an alternative to what I see posted so far. I think. Choose one special corner $A$. Note that $A$ has three adjacent vertices $B_1,B_2,B_3$.
Any permutation of the cube maps $A$ to one of the corners. So there are $8$ possibilities for where to send $A$ to.
Once $A$ is repositioned to $\varphi(A)$, you need to decide how $B_1,B_2,B_3$ will map to the three neighbors of $\varphi(A)$. There are $3!=6$ ways to do this and they are all valid. So now there are $8\cdot6=48$ possibilities.
At this point you have no flexibility for where the remaining four vertices get mapped to. Each one is the antipode for one of $A,B_1,B_2,B_3$ and so must land at the corresponding antipode to $\varphi(A),\varphi(B_1),\varphi(B_2),\varphi(B_3)$. So there are $48$ symmetries total and that's it.
Now when you think back, of the $6$ ways to relocate $B_1,B_2,B_3$, half are orientation preserving and half are not. Thus you get the $24$ rotations and the other $24$ symmetries that cannot be obtained through 3D spatial rotation alone. (Some of those are like what you might see if you held the cube up to a mirror.)
Also, each one of $A,B_1,B_2,B_3$ starts on a different diagonal. And the same is true of $\varphi(A),\varphi(B_1),\varphi(B_2),\varphi(B_3)$. So another way to look back on this is that we permuted the four corresponding diagonals $d,d_1,d_2,d_3$, coupled with a choice for which end of $\varphi(d_1)$ on which $\varphi(A)$ will land. Once all that is decided, the exact location of $\varphi(B_1),\varphi(B_2),\varphi(B_3)$ is determined. For example, $\varphi(B_1)$ is on the diagonal $\varphi(d_1)$, but only one of its two endpoints is adjacent to $\varphi(A)$. That comes to $4!\cdot2=48$ as well.