Explanation:

{
  1≥|⊃⍵-(+/÷≢)⍵: 0        ⍝ Base case: If it's the last iteration, return 0
         ⍵⌹=⍨⍵           ⍝ Average of ⍵ (the array) (not sure why that works)
            =⍨⍵           ⍝ Compare ⍵ to itself to create an array of 1s the same size as ⍵
         ⍵⌹               ⍝ ⍵ divided by that (matrix division)
       ⍵-                  ⍝ Subtract that from all elements of ⍵
      ⊃                    ⍝ Take only the first of those differences
     |                     ⍝ Absolute value
   1≥                      ⍝ Is it less than or equal to 1?
⋄1+∇h,3÷⍨3+/⍵,h←⊃⍵
               h←⊃⍵        ⍝ Assign the first element of ⍵ to h
            ⍵,             ⍝ Append to ⍵ (because of wrapping)
         3+/               ⍝ Take groups of 3 adjacent elements and sum each
      3÷⍨                  ⍝ Divide each sum by 3 (to get average)
    h,                     ⍝ Prepend h, which stays constant
   ∇                       ⍝ Call on this new iteration
 1+                        ⍝ Add 1 to that
}

Adám's idea of using ⌹ to find the average is really cool. See their explanation in the comments. R is the result of X⌹Y, and it's chosen to minimize the square of the difference between X and the matrix product of R and Y. In this case, since Y is a vector of all 1s, the average of X minimizes that squared difference.

Explanation:

{
  1≥|⊃⍵-(+/÷≢)⍵: 0        ⍝ Base case: If it's the last iteration, return 0
         ⍵⌹=⍨⍵           ⍝ Average of ⍵ (the array)
            =⍨⍵           ⍝ Compare ⍵ to itself to create an array of 1s the same size as ⍵
         ⍵⌹               ⍝ ⍵ divided by that (matrix division)
       ⍵-                  ⍝ Subtract that from all elements of ⍵
      ⊃                    ⍝ Take only the first of those differences
     |                     ⍝ Absolute value
   1≥                      ⍝ Is it less than or equal to 1?
⋄1+∇h,3÷⍨3+/⍵,h←⊃⍵
               h←⊃⍵        ⍝ Assign the first element of ⍵ to h
            ⍵,             ⍝ Append to ⍵ (because of wrapping)
         3+/               ⍝ Take groups of 3 adjacent elements and sum each
      3÷⍨                  ⍝ Divide each sum by 3 (to get average)
    h,                     ⍝ Prepend h, which stays constant
   ∇                       ⍝ Call on this new iteration
 1+                        ⍝ Add 1 to that
}

APL (Dyalog Unicode), 35 bytes

{1≥|⊃⍵-(+/÷≢)⍵:0⋄1+∇h,3÷⍨3+/⍵,h←⊃⍵}

Try it online!

A recursive function.

Explanation:

{
  1≥|⊃⍵-(+/÷≢)⍵: 0        ⍝ Base case: If it's the last iteration, return 0
         (+/÷≢)⍵           ⍝ Average of ⍵ (the array)
          +/               ⍝ Sum
            ÷              ⍝ Divided by  
              ≢             ⍝ Length
       ⍵-                  ⍝ Subtract that from all elements of ⍵
      ⊃                    ⍝ Take only the first of those differences
     |                     ⍝ Absolute value
   1≥                      ⍝ Is it less than or equal to 1?
⋄1+∇h,3÷⍨3+/⍵,h←⊃⍵
               h←⊃⍵        ⍝ Assign the first element of ⍵ to h
            ⍵,             ⍝ Append to ⍵ (because of wrapping)
         3+/               ⍝ Take groups of 3 adjacent elements and sum each
      3÷⍨                  ⍝ Divide each sum by 3 (to get average)
    h,                     ⍝ Prepend h, which stays constant
   ∇                       ⍝ Call on this new iteration
 1+                        ⍝ Add 1 to that
}

APL (Dyalog Unicode), 35 33 bytes

Saved 2 bytes thanks to Adám (and would have saved 1 with a tip from Bubbler)

{1≥|⊃⍵-⍵⌹=⍨⍵:0⋄1+∇h,3÷⍨3+/⍵,h←⊃⍵}

Try it online!

A recursive function.

Explanation:

{
  1≥|⊃⍵-(+/÷≢)⍵: 0        ⍝ Base case: If it's the last iteration, return 0
         ⍵⌹=⍨⍵           ⍝ Average of ⍵ (the array) (not sure why that works)
            =⍨⍵           ⍝ Compare ⍵ to itself to create an array of 1s the same size as ⍵
         ⍵⌹               ⍝ ⍵ divided by that (matrix division)
       ⍵-                  ⍝ Subtract that from all elements of ⍵
      ⊃                    ⍝ Take only the first of those differences
     |                     ⍝ Absolute value
   1≥                      ⍝ Is it less than or equal to 1?
⋄1+∇h,3÷⍨3+/⍵,h←⊃⍵
               h←⊃⍵        ⍝ Assign the first element of ⍵ to h
            ⍵,             ⍝ Append to ⍵ (because of wrapping)
         3+/               ⍝ Take groups of 3 adjacent elements and sum each
      3÷⍨                  ⍝ Divide each sum by 3 (to get average)
    h,                     ⍝ Prepend h, which stays constant
   ∇                       ⍝ Call on this new iteration
 1+                        ⍝ Add 1 to that
}

APL (Dyalog Unicode), 35 bytes

{1≥|⊃⍵-(+/÷≢)⍵:0⋄1+∇h,3÷⍨3+/⍵,h←⊃⍵}

Try it online!

A recursive function.

APL (Dyalog Unicode), 35 bytes

{1≥|⊃⍵-(+/÷≢)⍵:0⋄1+∇h,3÷⍨3+/⍵,h←⊃⍵}

Try it online!

A recursive function.

Explanation:

{
  1≥|⊃⍵-(+/÷≢)⍵: 0        ⍝ Base case: If it's the last iteration, return 0
         (+/÷≢)⍵           ⍝ Average of ⍵ (the array)
          +/               ⍝ Sum
            ÷              ⍝ Divided by  
             ≢             ⍝ Length
       ⍵-                  ⍝ Subtract that from all elements of ⍵
      ⊃                    ⍝ Take only the first of those differences
     |                     ⍝ Absolute value
   1≥                      ⍝ Is it less than or equal to 1?
⋄1+∇h,3÷⍨3+/⍵,h←⊃⍵
               h←⊃⍵        ⍝ Assign the first element of ⍵ to h
            ⍵,             ⍝ Append to ⍵ (because of wrapping)
         3+/               ⍝ Take groups of 3 adjacent elements and sum each
      3÷⍨                  ⍝ Divide each sum by 3 (to get average)
    h,                     ⍝ Prepend h, which stays constant
   ∇                       ⍝ Call on this new iteration
 1+                        ⍝ Add 1 to that
}