Field Extensions and Number of Isomorphisms 
The picture above is from Dummit and Foote, Third Edition. Later in the book, we find    

Clearly, the condition of equality is not necessary, as seen by taking the polynomial $ f(x) =  ( x ^{2} - 2) ( x ^{2} - 2)  $, which is not separable, but its splitting field, $    \mathbb{Q}     ( \sqrt{2   }     )          $ also has two automorphisms, and the degree of extension over $ \mathbb{Q} $ is also $ 2 $.    
However, in the book of Abstract Algebra by the author John A. Beachy, Third Edition, we find that    

which disagrees with the definition given by Dummit and Foote. I realize that the terminology of Beachy is, according to Wikipedia, old fashioned, but I redid the proof of Proposition 5 with the "old" definition, and it seems that the condition of equality becomes necessary too. 
Question. If we use Beachy's definition, does the equality condition in Proposition 5 become if and only if?       
 A: Everyone agrees that an irreducible polynomial over a field is separable iff it has distinct roots in an algebraic closure.  There is a well-known (standard?!?) lack of agreement as to when to call a reducible polynomial separable.  
I prefer Dummit-Foote's definition, though I wish it stated explicitly that the roots are living in an algebraic closure.  As in @Bernard's comment, this definition renders the Derivative Criterion valid, and (thus) it is faithfully preserved by base extension: in plainer language that means that if $f \in F[t]$ and $K/F$ is any field extension, then $f$ is separable when regarded as a polynomial over $F$ iff $f$ is separable when regarded as a polynomial over $K$.  In Beachy-Blair's definition, this does not work upon passing from any imperfect field to its algebraic closure.
Having said that: I'm not sure I understand your complaint about the proposition in Dummit-Foote.  Is it that in the given case the polynomial is not separable according to their definition so the result doesn't apply, but if you used Beachy-Blair's definition you could apply the result?  This seems a bit superficial: the result is actually true whenever the finite extension $E/F$ is normal (i.e., the splitting field of some polynomial) and separable, as is a standard result in most field theory texts.  If you want to concentrate on polynomials, you can just observe that the splitting field of a polynomial depends only on the largest squarefree polynomial divisor: in other words, if $f = \prod_{i=1}^r p_i^{e_i}$ with each $p_i$ irrreducible, then the splitting field of $f$ is certainly equal to the splitting field of $\prod_{i=1}^r p_i$.
Ah, I found your question.  The answer is yes.  
