Do there exist positive definite matrices $A$ and $B$ satisfying this condition? Denote by $\mathbb{P}_n$ be the set of real symmetric positive definite $n \times n$ matrices. In $\mathbb{P}_n$, we define an partial order as follow: for $X, Y \in \mathbb{P}_n$, we say $X \prec Y$ if $Y-X$ is also a positive definite matrix. Let $\lambda = 2^{1-1/p}$ with $p>1$ be a real number.
Question: For $0\prec X \prec Y \prec \lambda X$, do there exist positive definite matrices $A, B$ such that
$$X = \dfrac{A+B}{2}, \qquad Y = \left(\dfrac{A^p+B^p}{2}\right)^{1/p}?$$

Remark: The answer in the case of a scalar is Yes, and $\lambda = 2^{1-1/p}$ comes from this case to ensure that the function $$f(a) = \left(\dfrac{a^p + (2x-a)^p}{2}\right)^{1/p}$$ is surjective from $[0,2x]$ to $[x, \lambda x]$.
 A: Edit: The original answer only dealt with $p=2$, now the case $1<p<\infty$ is also dealt with. I leave the $p=2$ case as it is simpler.
No. A positive answer to your question for $p=2$ would in particular imply that
$$ 0<X<Y<\lambda X \implies X^2 \leq Y^2,$$
which is well-know to be false (the square map is not operator monotone), see below for an explicit counterexample.
Indeed, if
$$X = \dfrac{A+B}{2}, \qquad Y = \left(\dfrac{A^2+B^2}{2}\right)^{1/2},$$
then we have $Y^2 - X^2 = \dfrac{(A-B)^2}{4}$.
For a concrete counterexample, take $X=\begin{pmatrix} 1 &0\\0& \frac 1 4\end{pmatrix}$ and $Y = X+ t \begin{pmatrix} 1&1\\1&1\end{pmatrix}$. For small positive $t$, the inequalities $0<X<Y<\lambda X$ are satisfied, but $Y^2-X^2$ has determinant $- (3t/4)^2<0$, so is never positive.

When $1<p<\infty$, we can apply a similar argument. We shall use the following theorem by Loewner (I think): the map $t \mapsto t^r$ is operator concave if $0<r \leq 1$, and operator convex for $1\leq r \leq 2$. See for example Bhatia's book on matrix analysis.
So if we take some $1<r \leq \min(2,p)$, we have for every positive $A,B$, and
$$X = \dfrac{A+B}{2}, \qquad Y = \left(\dfrac{A^p+B^p}{2}\right)^{1/p},$$
$$ X^r \leq \frac{A^r+B^r}{2} \leq \left(\frac{A^p+B^p}{2}\right)^{r/p} = Y^{r}.$$
(the first inequality is the operator convexity of $t^r$, the second is the operator concavity of $t^{r/p}$.
So a positive answer to your question would imply that
$$ 0<X<Y<\lambda X \implies X^r \leq Y^r,$$
which should contradict the non-operator monotonicity of $t^r$. Something needs to be checked when $r<2$ as the non-operator monotonicity of $t^r$ only says that $0<X<Y \implies X^r \leq Y^r$ is not true, but I guess (did not check) that the same concrete counterexample as for $r=2$ above should work.
