How many positive integers $n$ are there such that $n$ is a multiple of $5$, and the least common multiple of $5!$ and $n$ equals $5$ times the greatest common divisor of $10!$ and $n$?
This is equivalent to counting the number of solutions to the following equation where $5\mid n$:
$$lcm(5!, n) = 5\gcd(10!, n)\Leftrightarrow lcm(2^3\times 3\times 5, n) = 5\gcd(2^8\times 3^4\times 5^2\times 7, n)$$
Now it is clear that the prime factors of $n$ can only consist of $2$, $3$, $5$ and $7$ because the right side of the above relation does not have any other prime numbers. Furthermore, it can must have between $3$ and $8$ twos, $1$ and $4$ threes and $0$ or $1$ seven. For five, because $n$ is a multiple of $5$ and the right side is multiplied by $5$, it will force $n$ can only only $3$ fives.
Then, by the multiplication principle, the total number of possibilities must be $$6\times 4\times 2\times 1=\boxed{48}$$