If a distribution’s scale parameter cannot equal 1, is it part of a scale family? In general if $f$ is a scale family we have that if $X\sim f(x\mid\lambda)$ then 
$\frac{X}{\lambda}\sim f(x\mid 1)$.
However what if $f$ has the constraint that its scale parameter $\lambda \in (1, \infty ).$ So scale cannot be $1.$
Now what is the distribution of $\frac{X}{\lambda}$?  We would think it would be $f(x|1)$ with scale $1,$ but as mentioned before the scale cannot equal 1.  
(Updated to address the comments - I think my original Weibull example did not add much to the question, so just sticking with general $f$. However, I had originally made $f$ a Weibull and constrained the scale parameter as above, just for example. I include this just because it is mentioned in the accepted answer below.)
Why constrain the scale? I am not sure - I just found a problem where it asked to prove that $f(x|\lambda)$ is a distribution, but added the constraint above.  Using a location scale theorem, we can prove this by just showing that $f(x|1)$ is a distribution, which is sometimes easier.  However, can we still do this if the constraint is present?
 A: The Weibull distribution $f(x \vert k,\lambda) = k x^{k-1} \lambda^{-k} e^{-(x/\lambda)^k}$ belongs to the scale family. Because when the space of parameters is $\lambda,k \in (0,\infty)$, then for any variable $X$ that is distributed as this Weibull distribution then some scaled version of this variable $c \cdot X$, for any value $c \in (0,\infty)$ will also be Weibull distributed (and more in particular the Weibull distribution of the scaled variable $c \cdot X$ will be just relating to a scaling of the parameter $\lambda$). 
If for whatever reason you consider an altered, more restricted, version of this Weibull distribution where the scaling parameter $\lambda$ is not in the range $(0,\infty)$, then you will not have a distribution that is in the scale family anymore. That is because this restricted Weibull distribution is not anymore closed under scaling, which is a requirement for being distribution that is part of the scale family of probability distributions.
Say you consider $X \sim f(x \vert k=1, \lambda=2)$ then $X/2 \sim f(x \vert k=1, \lambda=1)$, but when you restrict the set of distributions such that $f(x \vert k=1, \lambda=1)$ aint in it then this means that this restricted set of distributions is not part of the scale family.
