Realizing proper pure octonions as conjugates Let us take the octonions as having all integer coefficients and the multiplication table at BAEZ
We have a standard conjugation operator with $\bar{1} = 1$ and $\bar{e_i}= - e_i,$ extend by linearity. Then we get a norm, or squared Euclidean length, as
$Nx = x \bar{x} = \bar{x} x,$ and the $N(xy) = Nx \; Ny.$ We get the real part $\mathcal Rx = \frac{1}{2} (x + \bar{x}). $ We will call an octonion $o$ pure if its real part is $0,$ that is $\bar{o} = -o.$ We also retrieve the ordinary dot product with $(x,y)= \mathcal R(x\bar{y}).$ Among a million other facts, multiplication on either the left or the right by a fixed octonion preserves angles, and multiplication by any of the $e_i$ rotates everything $90^\circ.$
The main law we need is the  "alternative algebra" law,
$$  (xy)x = x(yx).  $$ I like the quick summary of the laws in
Robert Wilson 2008 SEMINAR PDF 
LEMMA:  $$ (\bar{t} x) t =   \bar{t} (x t)  $$
Proof: Write $t = r + o $ with $r$ real and $o$ pure. The $\bar{t} = r - o.$ Calculation gives
$$  (\bar{t} x) t = r^2 x + r x o - r o x - (ox)o,  $$
while
$$  \bar{t} (x t) = r^2 x + r x o - r o x - o(xo).  $$
Since $ (ox)o = o (xo)   $ the lemma follows.
To the question of interest, which was done for the quaternions as  the matrix in formula (10) on page 755 of Pall Automorphs (1940) and Theorem 3 on page 176 of Jones and Pall (1939), both available as pdfs at 
TERNARY 
Let 
$$   o = o_1 e_1 +   o_2 e_2 +  o_3 e_3 +  o_4 e_4 +  o_5 e_5 +  o_6 e_6 +  o_7 e_7           $$
be a "proper" pure quaternion, that is $\gcd(o_1,o_2,o_3,o_4,o_5,o_6,o_7) = 1,$ with square norm
$$  No = m^2 = o_1^2 +  o_2^2 +  o_3^2 +  o_4^2 +  o_5^2 +  o_6^2 +  o_7^2    $$ for integer $m.$ 
QUESTION: is it necessarily the case that we can find some integral octonion $t,$ with $Nt = m,$ such that
$$  o = \pm \bar{t} (e_j t)    $$ for some $j =1,2,3,4,5,6,7?$
EDDITTT: I got an email from a kind fellow who has a Maple package for playing with octonions, who says that with $o = e_4 + e_5 + e_6 + e_7$ of norm 4 has no expression as $  o =  \bar{t} (e_1 t).    $ I have asked him if checking all possible $\pm e_j$ makes any difference. It is a big help to be able to experiment, I don't have any computer programs myself this time. I will put in his name if he says that is alright. Oh, his Maple program may be using a slightly different multiplication table from the one I picked, i think there are 240 or 480 possible versions. 
Note that the proof by Jones and Pall uses associativity and, essentially, unique factorization (Theorem 2) for a proof by induction on the number of prime factors in out $m.$ That proof cannot work in quite such a pleasant manner. That is, their Theorem 2 is likely false, but perhaps this (Theorem 3) can be recovered as a very special case. Note that I can live with $m$ being odd, Jones and Pall needed to do that. For seven squares I do not think it should be necessary.
Oh, automorphs. It is fairly likely that every rational automorph of the sum of seven squares can be written with the seven rows being $\bar{t} (e_i t),$ and then divide by $m.$ I'm just saying.
EDIT, JUNE 7: note that the question to which I was  trying to apply octonions has an affirmative answer due to J. S. Hsia, see Filling in a rational orthogonal matrix given one row  while the octonion approach to seven variables was just not very effective. That is, it works for some 7-tuples but the percentage of failures goes up as the coefficients increase in size. 
 A: I'm not sure whether you were exploring a particular application of the octonions, but I want to point out that understanding the spaces where $o=\pm \bar{t} ( e_j t )$ exists are interesting. Papers like http://arxiv.org/abs/1002.1497 or http://arxiv.org/abs/0911.2255 use such constructs, and you're wondering about the further restriction of $t$ and $o$ being a 'proper octonion'. Thanks for posting your question, and some investigation. I wonder what actually are the spaces where a $t$ exists for some $e_j$ such that $o=\pm \bar{t} ( e_j t )$ for a given $o$? Quick question: With "integral octonion" did you mean any octonion that is made from integral coefficients? Or did you mean the (arguably) more customary use where "integral octonions" refers to a (reschaled) version of the E$_8$ lattice? Thanks, Jens
A: Sometimes nobody knows a question because it is just nonsense.
For distinct indices $i,j,k = 1,2,3,4,5,6,7,$ I have proved this little item by checking 35 triples in the Baez multiplication table:
$$  (e_i e_j) e_k = (e_j e_k) e_i = (e_k e_i) e_j.   $$ As a result,
$$   (e_i e_j) e_k  +  (e_k e_j) e_i = 0.  $$
I was pretty sure I had proved these with some calculations that forced the second version, but I also wanted to check, and it's true.  
W. Edwin Clark, emeritus at the University of South Florida, told me the example i edited into the question. A little more is true: if $Nt = 2,$ then the only possibilities I found for $\bar{t} e_i t$ are $\pm 2, \pm 2 e_j.$ 
Next I went ahead and checked $Nt = 3.$ This is also a complete bust. The  possibilities for $\bar{t} e_i t$ are $\pm 3, \pm 3 e_j, \; \pm e_j \pm 2 e_k \pm 2 e_m.$ There is simply not enough variety to create anything with $9 = 4 + 1 + 1 + 1 + 1 + 1 + 0.$
In conclusion, this was stupid, but now I know a bit more than I did.
A: I agree! 
The process of taking Norm or Conjugates of Octonions (as premised by the question) creates a trivial answer to the question. 
Just FYI, I too have a Mathematica notebook here that addresses octonions (and its Cayley-Dickson quaternion half). See that here. This is referenced in the mathematica.stackexchange.com forum here.
