I believe the answer can be found in OEIS. You have to add the paths of length $4$ through $9$ on a $3\times3$ grid, so $80+104+128+112+112+40=576$
I have validated the $80$, $4$ number paths. If we number the grid $$\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9 \end{array}$$
The paths starting $12$ are
$1236, 1254, 1258, 1256$
and there were $8$ choices of corner/direction, so $32$ paths start at a corner.
Starting at $2$, there are
$2145,2147,2369,2365,2541,2547,2587,2589,2563,2569$ for $10$ and there are $4$ edge cells, so $40$ start at an edge.
Starting at $5$, there are $8$ paths-four choices of first direction and two choices of which way to turn
Added per user3123's comment that cycles are allowed: unfortunately in OEIS there are a huge number of series titled "Number of n-step walks on square lattice" and "Number of walks on square lattice", and there is no specific definition to tell one from another. For $4$ steps, it adds $32$ more paths-four squares to go around, four places to start in each square, and two directions to cycle. So the $4$ step count goes up to $112$. For longer paths, the increase will be larger. But there still will not be too many.
Best Answer
Let us mark all the points as shown. C - Corner, M - Middle, T- Center.
Each C can reach 2 M OR 1 T, Each M can reach 2 M, 2 C OR 1 T, T can reach 4 C OR 4 M.
Step 1: 1 step password is allowed, so there are 9 ways.
Step 2: Out of the 9 ways of step 1, there are 4 Cs, 4 Ms and 1 T. Each of the 4 Cs can reach 2 Ms or 1 T, each of the 4 Ms can reach 2 Ms, 2 Cs or 1 T, 1 T can reach 4 Cs or 4 Ms. So there are $12+20+8=40$ ways to complete step 2.
Step 3: Collect all the Cs and Ms together. Now the choice of Ms for Ms will be reduced by 1, since we can't go back to from where we came. So $18+32+16=66$ ways to complete step 3.
Same way
$\underline{\text{Step 4}:} 62 $ways$\hspace{50 pt}$ $\underline{\text{Step 5}:} 62 $ways$\hspace{50 pt}$ $\underline{\text{Step 6}:} 62 $ways$\hspace{50 pt}$
$\underline{\text{Step 7}:} 62 $ways$\hspace{50 pt}$ $\underline{\text{Step 8}:} 62 $ways$\hspace{50 pt}$ $\underline{\text{Step 9}:} 62 $ways$\hspace{50 pt}$
So TOTAL number of combinations=No of ways of step 1+No of ways of step (1*2)+No of ways of step (1*2*3) +...+ No of ways of step (1*2*3*...*9)
NOTE: I was not aware about what is mentioned in the previous answer and comments that we can reach any point from any point.