= Solution
The equal-<variance> form of <Slepian's lemma> is the following. If $U,V$ are centered vectors with <multivariate normal distributions>, $\mathbb EU_i^2=\mathbb EV_i^2$ for every $i$, and $\mathbb EU_iU_j\geq\mathbb EV_iV_j$ whenever $i\ne j$, then
$$
\mathbb P(U_i\leq u_i\text{ for all }i)\geq\mathbb P(V_i\leq u_i\text{ for all }i)
$$
for every threshold vector $(u_i)$. In particular $\mathbb P(\max_iU_i\leq u)\geq\mathbb P(\max_iV_i\leq u)$. No nonsingularity of the <covariance matrices> is required.
Here the pointwise <variances> $2t^{2H}$ and $2t^{2H'}$ differ. To make the comparison rigorous, use the increment form, namely the <Sudakov-Fernique inequality>: if centered <multivariate normal distributions> satisfy $\mathbb E(U_i-U_j)^2\leq\mathbb E(V_i-V_j)^2$ for every pair, then $\mathbb E\max_iU_i\leq\mathbb E\max_iV_i$. Unlike the preceding form of <Slepian's lemma>, this statement has no equal-pointwise-<variance> hypothesis.
For completeness, the increment form follows by <Gaussian integration by parts>. Make $U,V$ <independent>, define $Z_r=\sqrt{1-r}U+\sqrt rV$ for $0<r<1$, and let
$$
F_\tau(z)=\frac1\tau\log\sum_{i=1}^n e^{\tau z_i},\qquad p_i(z)=\frac{e^{\tau z_i}}{\sum_j e^{\tau z_j}},\qquad D=\operatorname{Cov}(V)-\operatorname{Cov}(U).
$$
The $p_i$ are <softmax function> weights, with $p_i\geq0$ and $\sum_i p_i=1$. Direct differentiation gives $\partial_iF_\tau=p_i$ and $\partial_{ij}F_\tau=\tau(p_i\mathbf1_{i=j}-p_ip_j)$. The <Gaussian integration by parts> identity therefore yields
$$
\begin{aligned}
\frac d{dr}\mathbb EF_\tau(Z_r)&=\frac12\sum_{i,j}D_{ij}\mathbb E[\partial_{ij}F_\tau(Z_r)]\\
&=\frac\tau4\mathbb E\sum_{i,j}p_ip_j(D_{ii}+D_{jj}-2D_{ij})\geq0.
\end{aligned}
$$
The last summands are nonnegative because $D_{ii}+D_{jj}-2D_{ij}=\mathbb E(V_i-V_j)^2-\mathbb E(U_i-U_j)^2$. Since
$$
\max_i z_i\leq F_\tau(z)\leq\max_i z_i+\frac{\log n}{\tau},
$$
integration in $r$ and the limit $\tau\to\infty$ prove the <Sudakov-Fernique inequality>. All derivatives needed here are bounded for a fixed $\tau$. Singular <covariance matrices> can be handled by adding small <independent> centered <normal> noise with the same variance to both vectors, proving the identity for the nonsingular vectors, and letting the noise variance decrease to zero.
Use the printed normalization of <fractional Brownian motion>, which omits the conventional factor $1/2$. Its increment <variance> is
$$
\mathbb E|X_H(t)-X_H(s)|^2=2|t-s|^{2H},\qquad \mathbb E|Y_{H'}(t)-Y_{H'}(s)|^2=2|t-s|^{2H'}.
$$
For $s,t\in[0,1]$ and $H>H'$, $|t-s|^{2H}\leq|t-s|^{2H'}$. Enumerate $T$ by increasing finite subsets $T_n$ whose union is $T$, with $0\in T_n$. The <Sudakov-Fernique inequality> gives
$$
\mathbb E\max_{t\in T_n}X_H(t)\leq\mathbb E\max_{t\in T_n}Y_{H'}(t).
$$
Both <Gaussian processes> vanish at zero almost surely. Thus these maxima are nonnegative and increase with $n$; <monotone convergence theorem> gives
$$
\mathbb E\sup_{t\in T}X_H(t)\leq\mathbb E\sup_{t\in T}Y_{H'}(t).
$$
Their <continuous modifications> agree with the original <Gaussian processes> simultaneously on the countable set $T$, after discarding a single null event. Continuity and density of the rational parameters make each rational <supremum> equal to the supremum over $[0,1]$ for these modifications.
The expected suprema are finite. Indeed, the canonical distance of $Y_{H'}$ is $d(s,t)=\sqrt2|t-s|^{H'}$. A Euclidean grid supplies a closed-ball <metric covering number> bound $N(\varepsilon)\leq1+\lceil(\sqrt2/\varepsilon)^{1/H'}\rceil$ for $0<\varepsilon\leq\sqrt2$. The <Dudley entropy integral> bound for an anchored centered separable <Gaussian process> is
$$
\mathbb E\sup_tY_{H'}(t)\leq C\int_0^{\sqrt2}\sqrt{\log N(\varepsilon)}\,d\varepsilon<\infty.
$$
The integral is finite since its integrand grows at most like a constant times $\sqrt{1+\log(1/\varepsilon)}$ near zero. The same argument applies to $X_H$.
For any real path vanishing at zero,
$$
\sup_{t\in T}|x(t)|\leq\sup_{t\in T}x(t)+\sup_{t\in T}(-x(t)).
$$
A centered <Gaussian process> has the same <probability law> as its negative. Combining this symmetry with the increment comparison gives the stronger bound
$$
\mathbb E\sup_{t\in T}|X_H(t)|\leq2\mathbb E\sup_{t\in T}X_H(t)\leq2\mathbb E\sup_{t\in T}Y_{H'}(t)\leq2\mathbb E\sup_{t\in T}|Y_{H'}(t)|.
$$
In particular the requested comparison is
$$
\boxed{\mathbb E\sup_{t\in T}|X_H(t)|\leq4\mathbb E\sup_{t\in T}|Y_{H'}(t)|.}
$$
The common normalization factor $\sqrt2$ affects both sides equally. The comparison is valid for every $0<H'<H<1$, including the equality cases $s=t$ or $|s-t|=1$ in the increment bounds.
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