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Haskell, 90 75 bytes

n(l)|d<-[[0|n<-l,n==a]|a<-l]=and[[0]<d!!0,all(==d!!0)d]--an!((())),,,0<[]||

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Each character appears 5 times.

For reference: old version, 90 bytes, each char 3 times:

a x|h:u<-[sum[1|d<-x,not(d/=c)]|c<-x],"  \"\\&,../1::>acdlmmnosst">[]=h>1&&all(not.(/=h))u

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Haskell, 90 75 72 bytes

a[i]|d:n<-[[i|n<-i,n==a]|a<-i]=and[[i]<d,[d|i<-n]==n]--aadd,,,,:::::<=||

Each character appears 6 times. The input string is taken as a singleton list.

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For reference, old versions:

75 bytes, each char 5 times

n(l)|d<-[[0|n<-l,n==a]|a<-l]=and[[0]<d!!0,all(==d!!0)d]--an!((())),,,0<[]||

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90 bytes, each char 3 times:

a x|h:u<-[sum[1|d<-x,not(d/=c)]|c<-x],"  \"\\&,../1::>acdlmmnosst">[]=h>1&&all(not.(/=h))u

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Haskell, 90 bytes

a x|h:u<-[sum[1|d<-x,not(d/=c)]|c<-x],"  \"\\&,../1::>acdlmmnosst">[]=h>1&&all(not.(/=h))u

Each char appears three times. I've used the string " \"\\&,../1::>acdlmmnosst" to fill up the missing characters.

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                        c<-x      -- for all char c in the input string x
     [1|d<-x,not(d/=c)]           -- take a 1 for every appearance of c in x
  sum                             -- add those 1s
h:u                               -- let h be the first sum and u the list
                                  -- of all other sums
     " ... ">[]                   -- a guard that always evaluates to true, but
                                  -- is used to fill up the missing characters
   = h>1 &&                       -- return True if h > 1 and
          all(not.(/=h))u         -- all the other sums are equal to h

Haskell, 90 75 bytes

n(l)|d<-[[0|n<-l,n==a]|a<-l]=and[[0]<d!!0,all(==d!!0)d]--an!((())),,,0<[]||

Try it online!

Each character appears 5 times.

For reference: old version, 90 bytes, each char 3 times:

a x|h:u<-[sum[1|d<-x,not(d/=c)]|c<-x],"  \"\\&,../1::>acdlmmnosst">[]=h>1&&all(not.(/=h))u

Try it online!

Haskell, 90 bytes

a x|h:u<-[sum[1|d<-x,not(d/=c)]|c<-x],"  \"\\&,../1::>acdlmmnosst">[]=h>1&&all(not.(/=h))u

Each char appears three times. I've used the string " \"\\&,../1::>acdlmmnosst" to fill up the missing characters.

Try it online!

Haskell, 90 bytes

a x|h:u<-[sum[1|d<-x,not(d/=c)]|c<-x],"  \"\\&,../1::>acdlmmnosst">[]=h>1&&all(not.(/=h))u

Each char appears three times. I've used the string " \"\\&,../1::>acdlmmnosst" to fill up the missing characters.

Try it online!

                        c<-x      -- for all char c in the input string x
     [1|d<-x,not(d/=c)]           -- take a 1 for every appearance of c in x
  sum                             -- add those 1s
h:u                               -- let h be the first sum and u the list
                                  -- of all other sums
     " ... ">[]                   -- a guard that always evaluates to true, but
                                  -- is used to fill up the missing characters
   = h>1 &&                       -- return True if h > 1 and
          all(not.(/=h))u         -- all the other sums are equal to h