Let $(X, \mathcal{B}, \mu)$ be a probability space and $f:X\rightarrow X$ be an ergodic dynamical system with respect to $\mu$. Let $A : X \rightarrow GL(d,\mathbb{R})$ be measurable and let
\begin{align*} \Phi : \mathbb{N}_0 \times X &\rightarrow GL(d,\mathbb{R})\\ (n, x) &\mapsto \Phi(n,x) := A(f^{n-1}(x))\cdots A(f(x))A(x) \end{align*} i.e. $(f, \Phi)$ forms a linear cocycle. Then the multiplicative ergodic theorem states that there exist $p\leq d$ real constant Lyapunov exponents $\lambda_p<\ldots <\lambda_1$ and a (random) filtration of $\mathbb{R}^d$
$$\{0\}\subset V_p(x)\subset \cdots \subset V_1(x) = \mathbb{R}^d$$
such that for all $v\in \mathbb{R}^d\backslash \{0\}$ the Lyapunov characteristic exponent $\lambda(x,v)$ equates to
$$\lambda(x, v) := \lim_{n\to\infty}\frac{1}{n}\log\|\Phi(n, x)v\| = \lambda_i \Longleftrightarrow v \in V_i(x) \backslash V_{i+1}(x)$$
My question is the following: Does there exist a criterion such that for all fixed $v\in \mathbb{R}^d$, $\lambda(\cdot, v) := x \mapsto \lambda(x, v)$ is constant $\mu$-almost everywhere? This seems to be true in a lot of cases.
Note that $\lambda(f(x), v) = \lambda(x, A(x)v)$ (as $V_i(f(x)) = A(x) V_i(x)$) but $\lambda(\cdot, v)$ is not in general $f$-invariant.
Additional note: The intuition is as follows: with respect to the Lebesgue measure $\rho$, $\rho$-almost every $v\in \mathbb{S}^{d-1}$, $\mu$-almost every $x\in X$, $\lambda(v, x) = \lambda_1$ because of the above. If for $v \in \mathbb{S}^{d-1}$, if we do not have $\lambda(v, x) = \lambda_1$ for $\mu$-almost every $x\in X$. Then it seems most likely that $\lambda(v, x) = \lambda_i$ for $\mu$-almost every $x\in X$ and some $i>1$.
This is related to this question: Proof for ergodic system, the Lyapunov exponents are constant almost everywhere
