This cannot happen. When the number of unique eigenvalues changes we can have a discontinuity in the eigenspaces. For example, let $$f(t)=\left\lbrace\begin{array}{cc} 0 & \text{for }t<0 \\ t & \text{for }0\leq t \leq 1 \\ 1 & \text{for } t>1 \end{array}\right.$$ and define the matrix $$A(t)=\left(\begin{array}{cc} 1 & f(t-1) \\ f(t-1) & f(t) \end{array}\right).$$ $A(t)$ has eigenvalues $t,1$ and eigenvectors $(0,1),(1,0)$ for $t \in [0,1]$. For $t \in [1,2]$ $A(t)$ has eigenvalues $t,2-t$ and eigenvectors $(1,-1),(1,1)$. There is no way for this continuous family of Hermitian matrices to have a continuous choice of eigenvectors as the eigenspaces make a 45° jump at $t=1$.

This cannot happen in general. When the number of unique eigenvalues changes we can have a discontinuity in the eigenspaces. For example, let $$f(t)=\left\lbrace\begin{array}{cc} 0 & \text{for }t<0 \\ t & \text{for }0\leq t \leq 1 \\ 1 & \text{for } t>1 \end{array}\right.$$ and define the matrix $$A(t)=\left(\begin{array}{cc} 1 & f(t-1) \\ f(t-1) & f(t) \end{array}\right).$$ $A(t)$ has eigenvalues $t,1$ and eigenvectors $(0,1),(1,0)$ for $t \in [0,1]$. For $t \in [1,2]$ $A(t)$ has eigenvalues $t,2-t$ and eigenvectors $(1,-1),(1,1)$. There is no way for this continuous family of Hermitian matrices to have a continuous choice of eigenvectors as the eigenspaces make a 45° jump at $t=1$.