JavaScript (Node.js),  69 63 57  54 bytes

s=>Buffer(s).map(c=>s|c%4<1?s=c:c+c*90%320%34%24-8)+''

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How?

We process the input string $s$ character by character.

We reuse $s$ as a flag: as soon as a numeric value is stored in it, we know that we must not update anything else.

To identify "p" (112) and "P" (80), we use the fact that their ASCII codes are multiples of $4$ and the ASCII codes of the other letters at the beginning of the string ("z", "Z" and vowels) are not.

To turn a vowel with ASCII code $c$ into its counterpart $n$ while leaving z and Z unchanged, we use the following function:

$$n=c+((((90times c) bmod 320)bmod 34)bmod 24)-8$$

 letter | ASCII code |  * 90 | % 320 | % 34 | % 24 | - 8 | new letter

--------+------------+-------+-------+------+------+-----+-----------------------

   'i'  |     105    |  9450 |  170  |   0  |   0  |  -8 | 105 -  8 =  97 -> 'a'

   'a'  |      97    |  8730 |   90  |  22  |  22  |  14 |  97 + 14 = 111 -> 'o'

   'o'  |     111    |  9990 |   70  |   2  |   2  |  -6 | 111 -  6 = 105 -> 'i'

   'z'  |     122    | 10980 |  100  |  32  |   8  |   0 | 122 +  0 = 122 -> 'z'

   'I'  |      73    |  6570 |  170  |   0  |   0  |  -8 |  73 -  8 =  65 -> 'A'

   'A'  |      65    |  5850 |   90  |  22  |  22  |  14 |  65 + 14 =  79 -> 'O'

   'O'  |      79    |  7110 |   70  |   2  |   2  |  -6 |  79 -  6 =  73 -> 'I'

   'Z'  |      90    |  8100 |  100  |  32  |   8  |   0 |  90 +  0 =  90 -> 'Z'

Commented

s =>                  // s = input string

  Buffer(s)           // convert it to a Buffer of ASCII codes

  .map(c =>           // for each ASCII code c in s:

    s |               //   if s is numeric

    c % 4 < 1 ?       //   or c is either 'p' or 'P':

      s = c           //     turn s into a numeric value and yield c

    :                 //   else:

      c +             //     update c

        c * 90 % 320  //     by applying the transformation function

        % 34 % 24     //     (see above)

        - 8           //

  ) + ''              // end of map(); coerce the Buffer back to a string

C (gcc),  81 ... 61 48  46 bytes

Saved 2 bytes thanks to @Grimy

Port of my JS answer. Outputs by modifying the input string.

f(char*s){for(;*++s%4;*s+=*s*90%320%34%24-8);}

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Commented

f(char * s) {       // f = function taking the input string s

  for(;             //   for each character *s in s:

    *++s % 4;       //     advance the pointer; exit if *s is either 'p' or 'P' (it's safe 

                    //     to skip the 1st character, as it's guaranteed to be 'z' or 'Z')

    *s +=           //     update the current character:

      *s * 90 % 320 //       apply a transformation formula that turns

      % 34 % 24     //       a vowel into the next vowel in the sequence

      - 8           //       while leaving 'z' and 'Z' unchanged

  );                //   end of for()

}                   // end of function

Stax, 19 bytes

Ç╛√êΣ%,╖FP╚`=Lh←⌡·ƒ

Run and debug it

Retina 0.8.2, 21 bytes

iT`Io`AOIiaoi`^.+?p

Try it online! Transliterates letters up to and including the first p, although the z and p aren't in the transliteration section so aren't affected. The first O is quoted because it normally expands to 13567 and the second o is quoted because it too is magic; in the first part of the transliteration it expands to the other string. The resulting transliteration is therefore from IAOIiaoi to AOIiaoi then removing the duplicate source letters results in IAOiao to AOIaoi.

C, 43 bytes

f(char*s){for(;*++s%4;*s^=*s%16*36%98%22);}

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Based on Arnauld's answer. I did a brute-force search to find the shortest formula that turns a => o, o => i, i => a, z => z.

Japt, 22 20 bytes

-2 bytes thanks to Oliver

r"%v+"_d`i¬iao¯`pu}"

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R, 110 76 bytes

-36 bytes thanks to Krill

This function takes an input of one string.

function(a)sub(s<-sub('z+(.+?)p.*','\1',a,T),chartr('aioAIO','oaiOAI',s),a)

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Perl 6, 41 33 bytes

{S:i{.+?p}=$/~~tr/iaoIAO/aoiAOI/}

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Simple case-insensitive substitution to shift the vowel section.

Python 3.8 (pre-release), 81 78 bytes

lambda s,p='iaoIAO':''.join(('aoiAOI'+c)[(p:=p*(ord(c)%4)).find(c)]for c in s)

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Python 2, 98 bytes

lambda s:''.join(('aoiAOI'+c)[(-~-('p'in s[:i].lower())*'iaoIAO').find(c)]for i,c in enumerate(s))

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Java (JDK), 52 bytes

s->{for(int i=0;s[++i]%4>0;)s[i]^=s[i]%16*36%98%22;}

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• Port of Grimy's C answer. Make sure to upvote his answer!

Perl 5, 31 bytes

s/.+?p/$&=~y,iaoIAO,aoiAOI,r/ei

TIO

SNOBOL4 (CSNOBOL4), 183 bytes

	INPUT (BREAK('Pp') SPAN('Pp')) . P REM . S

A	P 'a' ='o'	:S(A)

Z	P 'A' ='O'	:S(Z)F(P)

I	P 'i' ='a'	:S(I)

K	P 'I' ='A'	:S(K)F(P)

O	P 'o' ='i'	:S(O)

L	P 'O' ='I'	:S(L)

P	OUTPUT =P S

END

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Python 2, 84 bytes

f=lambda s,r='':s[0]in'pP'and r+s or f(s[1:],r+('aoiAOI'+s[0])['iaoIAO'.find(s[0])])

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C# (Visual C# Interactive Compiler), 60 bytes

n=>{for(int i=0;n[i]%4>0;)n[i]^=(char)(n[i++]%16*36%98%22);}

Based off of Grimy's C answer.

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Ruby, 40 bytes

$_[$&]=$&.tr"IAOiao","AOIaoi"if/z+.+?p/i

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05AB1E (legacy), 22 bytes

l'pkIg‚£ć…iaoDÀ‚Du+`‡ì

Try it online or verify all test cases.

Can definitely be golfed a bit more..

Uses the legacy version of 05AB1E instead of the Elixir rewrite, because + merges fields in same-length lists, whereas the new version would need a pair-zip-join ‚øJ instead.

Explanation:

l                       # Lowercase the (implicit) input-string

                        #  i.e. "ZipPeroni" → "zipperoni"

 'pk                   '# Get the index of the first "p"

                        #  i.e. "zipperoni" → 2

    Ig‚                 # Pair it with the length of the entire input-string

                        #  i.e. 2 and "zipperoni" → [2,9]

       £                # Split the (implicit) input-string into parts of that size

                        #  i.e. "ZipPeroni" and [2,9] → ["Zi","pPeroni"]

        ć               # Extract head: push the head and remainder separately to the stack

                        #  i.e. ["Zi","pPeroni"] → ["pPeroni"] and "Zi"

         …iao           # Push string "iao"

             DÀ         # Duplicate, and rotate it once towards the left: "aoi"

               ‚        # Pair them up: ["iao","aoi"]

                Du      # Duplicate and transform it to uppercase: ["IAO","AOI"]

                  +     # Python-style merge them together: ["iaoIAO","aoiAOI"]

                   `    # Push both strings to the stack

                    ‡   # Transliterate; replacing all characters at the same indices

                        #  i.e. "Zi", "iaoIAO" and "aoiAOI" → "Za"

                     ì  # Prepend it to the remainder (and output implicitly)

                        #  i.e. ["pPeroni"] and "Za" → ["ZapPeroni"]

PHP, 30 bytes

too simple for a TiO:

<?=strtr($argn,oiaOIA,iaoIAO);

Run as pipe with -nF. For PHP 7.2, put the string literals in quotes.

C/C++ (VC++ compiler) 192bytes

this is a rather naive try but anyway

void f(char*I){int c={-8,14,6},B=1,v={105,97,111},j=0;for(*I;*I>0&B;I++){if(*I==80|*I==112){B=0;break;}if(*I==90|*I==122){}else{for(j;j<3;j++){if(*I==v[j]|*I==v[j]-32){*I+=c[j];break;}}}}}

some slightly more readable Version is this

#include "stdafx.h"



void f(char * theString)

{

    signed int change = {'a'-'i','o'-'a','o'-'i'}; // add this to the vowel to get the next one

    char theVowels = {'i','a','o'};

    int breaker = 1;

    printf("Input %sn",theString);

    for (int i = 0;(theString[i] != '') && breaker; i++)

    {

        switch (theString[i])

        {

            case 'Z': /*fall through*/

            case 'z': break;

            case 'P': /*fall through*/

            case 'p': breaker = 0;

                      break; 

            default: 

            {

                for (int j = 0; j < 3; j++)

                {

                    if ((theString[i] == theVowels[j]) || (theString[i]==(theVowels[j]-'a'+'A')))

                    {

                        theString[i] += change[j];

                        break;

                    }

                }

            }

            break;

        }



    }

    printf("Output %sn",theString);

}

int main()

{

    char theString= "zzzzIIIIp0815-4711"; // a test string

    f(theString);

    return 0;

}