Ancient Caligraphy

Let's overview all digit changes upon turning them "upside down":

As it was mentioned, numbers 3, 4 and 7 cannot be used.

We are working with 6-digit integers. Let's see what is difference between the integer and its "upside down":

$\overline { { p }_{ 5 }{ p }_{ 4 }{ p }_{ 3 }{ p }_{ 2 }{ p }_{ 1 }{ p }_{ 0 } } -\overline { { q }_{ 0 }{ q }_{ 1 }{ q }_{ 2 }{ q }_{ 3 }{ q }_{ 4 }{ q }_{ 5 } } ={ 10 }^{ 5 }\cdot { p }_{ 5 }+{ 10 }^{ 4 }\cdot { p }_{ 4 }+{ 10 }^{ 3 }\cdot { p }_{ 3 }+{ 10 }^{ 2 }\cdot { p }_{ 2 }+{ 10 }^{ 1 }\cdot { p }_{ 1 }+{ 10 }^{ 0 }\cdot { p }_{ 0 }-{ 10 }^{ 5 }\cdot { q }_{ 0 }-{ 10 }^{ 4 }\cdot { p }_{ 1 }-{ 10 }^{ 3 }\cdot { p }_{ 2 }-{ 10 }^{ 2 }\cdot { p }_{ 3 }-{ 10 }^{ 1 }\cdot { p }_{ 4 }-{ 10 }^{ 0 }\cdot { p }_{ 5 }=[{ 10 }^{ 5 }\cdot({ p }_{ 5 }-{ q }_{ 0 })+{ 10 }^{ 0 }\cdot({ p }_{ 0 }-{ q }_{ 5 })]+[{ 10 }^{ 4 }\cdot({ p }_{ 4 }-{ q }_{ 1 })+{ 10 }^{ 1 }\cdot({ p }_{ 1 }-{ q }_{ 4 })]+[{ 10 }^{ 3 }\cdot({ p }_{ 3 }-{ q }_{ 2 })+{ 10 }^{ 2 }\cdot({ p }_{ 2 }-{ q }_{ 3 })]={ O }_{ 5, 0 }+{ O }_{ 4, 1 }+{ O }_{ 3, 2 }$

As we can see the difference is sum of three numbers. Notice that only ${ O }_{ 5, 0 }$ defines the last digit of the difference. Similarly the first digit is almost completely dependent on ${ O }_{ 5, 0 }$ , except it can be less by 1 because ${ O }_{ 4, 1 }$ or ${ p }_{ 0 }-{ q }_{ 5 }$ may be negative or other. That means:

$\begin{cases} { p }_{ 5 }-{ q }_{ 0 }=5\quad (or\quad 6) \\ { p }_{ 0 }-{ q }_{ 5 }=4\quad (or\quad -6) \end{cases}$

It's easy to see that the only possible digits are: ${ p }_{ 0 }=2, { p }_{ 5 }=8$ (correspondingly, ${ q }_{ 0 }=2, { q }_{ 5 }=8$ ). That means that ${ O }_{ 5, 0 }$ equals 599994. Thus ${ O }_{ 4, 1 }+{ O }_{ 3, 2 }=-79950$

Similarly digits -7 and 5 are defined by ${ O }_{ 4, 1 }$ , except the first digit can vary by 1 (plus or minus). Since ${ O }_{ 4, 1 }$ depends only on ${ p }_{ 1 }, { p }_{ 4 }$ , it's easy to find that the only possible values are ${ p }_{ 1 }=6, { p }_{ 4 }=1$ ( ${ q }_{ 1 }=9, { q }_{ 4 }=1$ ). That means ${ O }_{ 4, 1 }=-79950$ and ${ O }_{ 3, 2 }=0$ . Getting ${ O }_{ 3, 2 }=0$ is very easy because for every ${ p }_{ 2 }$ there is only ${ p }_{ 3 }$ that fits:

Thus we have $\boxed{7}$ possible integers that satisfy the condition: