Circular permutation with two iterators

I need to make an circular permutation of an list for example I have : (a,b,c,d,e) I want (e,a,b,c,d). But I don't succeed to do so, here what I tried :

#ifndef ALGORITHME_H
#define ALGORITHME_H

template<typename I>  
void permutationCirculaire(I first, I last) {
    typename std::iterator_traits<I>::value_type firstElement = *first;
    typename std::iterator_traits<I>::value_type res;
    I tmp = first;

    for (++first; tmp != last; ++tmp) {
        *first = *tmp;
        std::cout << "tmp : " << *tmp << ", first : " << *first << std::endl;
        ++first;
    }


}

#endif

I get this : tmp : a, first : a tmp : a, first : a tmp : a, first : a tmp : a, first : a tmp : a, first : a

And I don't know why, my main :

#include <iostream>
#include <list>
#include "algorithme.h"

using namespace std;

int main() {    
    list<char> liste;
    liste.push_back('a');
    liste.push_back('b');
    liste.push_back('c');
    liste.push_back('d');
    liste.push_back('e');

    cout << "( ";
    for (list<char>::iterator it = liste.begin(); it != liste.end(); ++it) {
        cout << *it << " ";
    }
    cout << ") " << endl;

    cout << "Permutation : " << endl;
    permutationCirculaire(liste.begin(),liste.end());

    cout << "( ";
    for (list<char>::iterator it = liste.begin(); it != liste.end(); ++it) {
        cout << *it << " ";
    }
    cout << ") " << endl;

    return 0; 
}

If you knwo why don't hesitate...

I finally succeed to correct my problem here is my solution, thanks everyone for your help :

template<typename I>  
void permutationCirculaire(I first, I last) {
    typename std::iterator_traits<I>::value_type current = *first;
    typename std::iterator_traits<I>::value_type save = *(--last);

    for(; first != last; ++first) {
        current = *first;
        *first = save;
        save = current;
    }
    *first = save;
}

Sorry again for the mistakes.