I don't like even

Without the digit $2,4,6,8$ , our sequence of natural numbers, in ascending order, will be

$1,3,5,7,9,10,11,13,15,17,19,30,...$

I will say this is a system in base $6$ . In this case, the base system is quite different. What I mean is after $6$ terms it gets into the next round.

If I let

$3\rightarrow2$

$5\rightarrow3$

$7\rightarrow4$

$9\rightarrow5$

Then the sequence become

$1,2,3,4,5,10,11,12,13,14,15,20,...$

and

$97531\rightarrow54321$

We will need to find the number $54321$ is in which term in this sequence. Hence,

$54321_6$ convert to base $10$ is $\boxed{7465_{10}}$ .

Let us consider the numbers that don't start with 9. The first digit can therefore be 7,5,3,1,0. The rest can be 9,7,5,3,1,0. One of these numbers created must be ignored, because
$00000\notin \left[ 1,97531 \right]$

Therefore, there are are $5\times { 6 }^{ 4 }-1.$

Now, let us consider the numbers that start with 9, but don't have 7 as their second digit. The first digit must be nine. The second digit can be 5,3,1,0. The rest of the digits can be any of the possible digits.

Therefore, there are $4\times { 6 }^{ 3 }$

Now, the numbers that start with 97, but don't have 5 as their third digit. After the nine seven, the third digit can therefore be 3,1,0. The fourth and fifth can be any of the 6 digits.

Therefore, there are $3\times { 6 }^{ 2 }$

You are probably beginning to see a pattern. Lovely, isn't it!

Now, the numbers beginning with 975, without a 3 as the fourth digit. That leaves 2 possibilities (1,0) for the fourth digit, and six for the fifth.

$2\times { 6 }^{ 1 }$

And then, 9753, without a 1 as the last digit. Just for the sake of the pattern, we will express this as

$1\times { 6 }^{ 0 }$

Finally, we add on our final number; 97531.

The answer, therefore, is $5\times { 6 }^{ 4 }+4\times { 6 }^{ 3 }+3\times { 6 }^{ 2 }+2\times { 6 }^{ 1 }+1\times { 6 }^{ 0 }=\quad 7465$

N.B. The +1 for the number 97531 is negated by the -1 for the number 00000

As the number doesn't contain the digits 2, 4, 6 or 8, the choice for each of its digits is 0, 1, 3, 5 or 9. This makes 6 choices for each digit and a maximum of 5 digits (if there are less, there will be zeros in front). So there are $6^{5}$ = 7776 numbers between 00000 and 99999 that use only the digits 1, 3, 5, 7, 9 or 0. Then we must subtract 1 because 00000 is too small. There are also a several possibilities where the number will be larger than 97531:

If the first two digits are 99, there are 6 x 6 x 6 = 216 possibilities for the last three digits, all of which will form a number bigger than 97531.

If the first three digits are 979, there are 36 possibilities which will all be too big.

977 gives another 6 x 6 = 36 possibilities.

9759, 9757 and 9755 give a further six numbers each. 6+6+6=18.

And, finally the 4 numbers 97539, 97537, 97535 and 97533 are all too big.

There are a total of 216+36+36+6+6+6+4=310 possibilities for numbers that are too large for the given criteria and 1 number that is too small.

So the final answer is $6^{5}$ - 310 - 1 = 7465 of which the last three digits are $\boxed{465}$ .