ecco aggiusto
Codice PHP:

<script language="javascript">

function showimage(){    
var image = document.getElementById("imageToSwap");    
var change = document.getElementById("colour");   
 image.src = change.value;   
};
</script>
<label for="esercizi">Esercizio: </label>
<br />
<select name="picture" size="1"  onChange="showimage()" id="colour">

    <?                   
   $dati
=mysql_query("SELECT * FROM esercizi WHERE id_gruppomuscolare = '$id_gruppo' ORDER BY nome");           while($array=mysql_fetch_array($dati))
{    
     echo 
"<option value=\"$array[immagine1]\">$array[nome]</option>";         
$red=$array[immagine1];          }                                 ?>             </select>
<br> 
 <img id="imageToSwap" src="<?php echo $red ?>
 name="pictures" width="150" border=0 alt="L'immagine appare qui">
</td></tr>