We start off by placing most of the $1$ and $2$ cells across the whole board, using all the given conditions to narrow down the possibilities. They can both be placed in the left-middle, centre, and right-middle boxes right away, and then we can make some more deductions in the upper and lower boxes:
Here the grey shaded cells are the ones which can be $3$, using the given condition 4. Notice that the $3$ in the left-middle box can only be in one of two possible places.
Now a slightly sneaky observation in the top-right box:
The $2$ cannot be in the top-middle cell, because then the $1$ couldn't be anywhere. So the $2$ must be one of the lower two possibilities, which means the $1$ can't be in the central cell, and must therefore be on the top row. That means we know where the $1$ is in the top-middle box, namely in the middle-left cell.
A similarly sneaky observation in the top-middle box:
The $2$ must be in either the top-middle cell, the top-right cell, or the middle-right cell. But putting it in the middle-right cell would leave no possibilities for $2$ in the top-right box, so it must be in the top row. Now we can place the $2$ in the top-left and top-right boxes.
Looking again at $3$ in the middle-left box:
if it's on the bottom, we get this with no possible position for $3$ in the top-right box. So $3$ is in the top-middle cell of the middle-left box, and we can immediately reduce the reminaing possibilities for placing $3$ by a LOT:
Now there's only one place for $3$ in the seventh row. Also, in the bottom-middle box,
if the $3$ is on the right-hand side, then it can't be anywhere in the bottom-right box, contradiction. Now we can place ALL the remaining $1$, $2$, and $3$ cells, and shade in the possibilities for $4$:
The fifth-row $4$ must be in the centre box, and can't be on the left as that would leave no possibilities for the middle-left box. Then we can fill in all the $4$ cells and shade possibilities for $5$:
Starting from the middle-right box, it's easy to fill in all the $5$ cells. Shading possibilities for $6$:
Starting from the centre and top-right boxes, it's easy to fill in all the $6$ cells. Then just keep going similarly with the $7$, $8$, and $9$ cells to get the final solution.