High Five!

By direct observation, all numbers that are made exclusively from odd digits and divisible by $5^5$ between $5$ and $6$ digits long all end in $59375$ .

Also, since $10^5 = 2^5 \cdot 5^5$ , this means that $10^5$ is divisible by $5^5$ , which means that the last $5$ digits of the multiples of $5^5$ repeat themselves in a cycle, so all numbers that are made exclusively from odd digits and divisible by $5^5$ for any number greater than $5$ digits long must all end in $59375$ with any combination of odd numbers in front of the $59375$ .

Therefore, since there are $5$ odd single digits ( $1$ , $3$ , $5$ , $7$ , and $9$ ), the number of numbers that are made exclusively from odd digits and divisible by $5^5$ for any number greater than $5$ digits long follow powers of $5$ , where the exponent is the number of spaces available in front of the $59375$ . In general, there are $5^{n - 5}$ numbers that can be made that are exactly $n$ digits long that are made exclusively from odd digits and divisible by $5^5$ .

Since the problem asks for $5^5$ numbers, $n - 5 = 5$ , and so $n = \boxed{10}$ .

When I tested 7-digit numbers using $Mathematica$

Select[FromDigits/@Tuples[{1,3,5,7,9},{7}],IntegerQ[#/5^5]&]

I got {1159375,1359375,1559375,1759375,1959375,3159375,3359375,3559375,3759375,3959375,5159375,5359375,5559375,5759375,5959375,7159375,7359375,7559375,7759375,7959375,9159375,9359375,9559375,9759375,9959375}

which are ALL possible numbers ending in 59375

then I checked that there are exactly $3125$ 5-tuples
Length@Tuples[{1, 3, 5, 7, 9}, {5}]

and I had the answer