Java 10, 291 287 233 229 bytes
n->{var t="";for(var d:n.split(t))t+=t.contains(d)?"":d;return p(n,"",t);}boolean p(String n,String p,String s){int l=s.length(),i=0;var r=n.contains(p);for(;i<l;)r&=p(n,p+s.charAt(i),s.substring(0,i)+s.substring(++i));return r;}
-4 bytes by taking inspiration from what @Arnauld mentioned in his JavaScript answer:
If all permutations of the \$N\$ symbols are present in the input string \$s\$, so are all prefixes of said permutations. Therefore, it's safe to test that all \$p\$ are found in \$s\$ even when \$p\$ is an incomplete permutation whose size is less than \$N\$.
That's why we can use a recursive function that recursively builds each permutation \$p\$ of the symbols and tests whether \$p\$ exists in \$s\$ at each iteration, even when \$p\$ is still incomplete.
Takes the integer-input as String.
Try it online.
Explanation:
n->{ // Method with String as parameter and boolean return-type
var t=""; // Temp String, starting empty
for(var d:n.split(t)) // Loop over the digits of the input:
t+= // Append to String `t`:
t.contains(d)? // If `t` contains this digit already:
"" // Append nothing
: // Else (it doesn't contain this digit yet):
d; // Append this digit
return p(n,"",t);} // Call the separated recursive method to check if each
// permutation of `t` is a substring of `n` and return it as
// Separated recursive method to get all permutations of String `t`, and check for each
// if it's a substring of String `n`
boolean p(String n,String p,String s){
int l=s.length(), // Get the length of the input-String `s`
i=0; // Set the index `i` to 0
var r= // Result-boolean, starting at:
n.contains(p); // Check that String `n` contains part `p` as substring instead
// (this doesn't necessary have to be the full permutation,
// but it doesn't matter if the part is smaller)
for(;i<l;) // Loop `i` in the range [0, length):
r&= // Add the following to the boolean-return (bitwise-AND style):
p( // Do a recursive call with:
n,p // The current part,
+s.charAt(i),// appended with the `i`'th character as new part
s.substring(0,i)+s.substring(++i));
// And the String minus this `i`'th character as new String
// (and increment `i` for the next iteration in the process)
return r;} // And return the resulting boolean
