I am not sure what you mean by "equisingular" stratification. But I guess you would like to say that *$X$ is equisingular along $Y$* if the local rings $\mathcal{O}_{X,x}$ have constant multiplicity for all $x \in Y$. The *equisingular stratification* would be a stratification with respect to the condition being *equisingular* along a subvariety.

In that case, equisingular stratifications and Whitney stratifications are not the same thing (though a Whitney stratification is always an equisingular stratification). You can construct a counter-example as follows.

First of all, if I remember correctly, if $Y \subset X$ is a smooth Whitney stratum, then $X$ is normally flat along $Y$. In particular, if $Y$ is closed in $X$, then the scheme-theoretic exceptional divisor of the blow-up of $X$ along $Y$ is flat over $Y$.

Now, let $Z$ be a generic hyperplane section of $\mathbb{P}^2 \times \mathbb{P}^1 \subset \mathbb{P}^5$. The surface $Z$ is a ruled cubic surface. I denote by $L$ the principal axis of the ruling.

Let us denote by $X \subset (\mathbb{P}^4)^*$ the projective dual of $Z$. It is easily checked that $X$ is a cubic hypersurface which is not a cone. In particular, all singular points of $X$ have exactly multiplicty $2$ in $X$.

Let $H$ be a hyperplane containing $L$. Since $H$ cuts $Z$ in a curve of degree $3$ (Bezout's Theorem), we have $H \cap Z = L \cup \mathcal{C}$, were $\mathcal{C}$ is a curve of degree $2$ on $Z$. In fact, one proves easily that $\mathcal{C}$ must be the union of two lines ruling $Z$, say $l_H$ and $l'_H$.

If $l_H \neq l'_H$, then $H$ is tangent to $Z$ along the two points $l_H \cap L$ and $l'_H \cap L$. Thus, $H^{\perp}$ is a singular point of $X$ and the tangency locus of $H$ along $Z$ is two distinct points.

If $l_H = l'_H$, then $H$ is tangent to $Z$ along $l_H$. Again $H$ is a singular point of $X$.

This shows that $L^{\perp}$ is in the singular locus of $X$. In fact, one proves that the singular locus of $X$ is exactly $L^{\perp}$. Hence, the singular locus of $X$ is a $\mathbb{P}^2$ (in particular it is smooth) and $X$ has multiplicity exactly $2$ along all the points of $H^{\perp}$. Hence $X - L^{\perp} ; L^{\perp}$ is an equi-singular stratification of $X$.

Let me prove that this stratification is not a Whitney stratification. By the preliminary remark, I only have to prove that $X$ is not normally flat along $L^{\perp}$.

Let $I_Z =\{(z,H^{\perp}) \in Z \times (\mathbb{P}^{N})^*, \textrm{such that} \, T_{Z,z} \subset H \}$ be the conormal variety of $Z$. Its projection on the second factor maps onto the projective dual of $Z$ (which is $X$).

If $Z$ is smooth and its projective dual is a hypersurface, it is a classical result of Nash that $I_Z$ is the blow-up of $X$ along its singular locus (which is $L^{\perp}$).

As a consequence, we see that the exceptional locus of this blow-up can not be flat over $L^{\perp}$. Indeed, if $p$ is the projection:

$$ p : I_Z \longrightarrow X$$

then the fiber of $p$ over $H^{\perp} \in L^{\perp}$ is:

_ two points when $l_H \neq l'_H$

_a line when $l_H = l'_H$.

This proves that $X$ is not normally flat along $L^{\perp}$ and so $X-L^{\perp}, L^{\perp}$ is not a Whitney stratification of $X$.

**Hope for Whitney = equisingular**. Teissier and his collaborators introduced in the 70's a much finer notion of multiplicity (which basically takes into account all the
multiplicities of the local rings of all the generic polars passing through a fixed point of $X$) and he proves in that context that Whitney = equisingular.
The paper where he proves this result is called "Variétés polaires II : Mulitplicités polaires, sections planes et conditions de Whitney". You can find it here : https://webusers.imj-prg.fr/~bernard.teissier/documents/VarPol2.pdf